Revision Notes for CBSE Class 11 Maths Chapter 12 (Limits and Derivatives) - Free PDF Download
Section–A (1 Mark Questions)
1. If $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$ and $n\epsilon N$ find n.
Ans. We have, $\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80$
$$\begin{aligned}& \Rightarrow n \cdot 2^{n-1}=80 \\& \Rightarrow n \cdot 2^{n-1}=5 \cdot 2^{5-1} \\& \Rightarrow n=5 .\end{aligned}$$
2. $\lim _{x \rightarrow 1} \frac{x^2+1}{x+100}=\frac{1^2+1}{1+100} =?$
Ans. We have,
$$\lim _{x \rightarrow 1} \frac{x^2+1}{x+100}=\frac{1^2+1}{1+100}=\frac{2}{101}$$
3. If $f(x)=x^n$ and $if f{}'(1)=10$ find the value of n.
Ans. We have, $f(x)=x^n$.
Differentiating both sides
w.r.t we get $f^{\prime}(x)=n x^{n-1}$.
Putting $x=1$ we get
$$f^{\prime}(1)=n \Rightarrow 10=n\left[\because f^{\prime}(1)=10\right]$$
4. Compute the derivative of $f(x)=6 x^{200}-x^{50}+x$.
Ans. We have, $f(x)=6 x^{200}-x^{50}+x$
$$\begin{aligned}\Rightarrow f^{\prime}(x) & =6\left(200 x^{190}\right)-\left(50x^{69}\right)+1 \\& =1200 x^{199}-50 x^{49}+1\end{aligned}$$
5. Evaluate $\lim _{x \rightarrow 0} \frac{x}{\cos x}$ .
Ans. $\lim _{x \rightarrow 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}=0$
Section–B (2 Marks Questions)
6. Evaluate: $\lim _{x \rightarrow 2} \frac{x^3-4 x^2+4 x}{x^2-4}$
Ans. Evaluating the function at 2 , it is of the form $\frac{0}{0}$,
Hence,
$$\lim _{x \rightarrow 2} \frac{x^3-4 x^2+4 x}{x^2-4}=\lim _{x \rightarrow 2}\frac{x(x-2)^2}{(x+2)(x-2)}$$
$$\begin{gathered}=\lim _{x \rightarrow 2} \frac{x(x-2)}{(x+2)} \text { as } x \neq 2 \\=\frac{2(2-2)}{2+2}=\frac{0}{4}=0 .\end{gathered}$$
7. Evaluate: $\lim _{x \rightarrow 2} \frac{x^3-2 x^2}{x^2-5 x+6}$
Ans. Evaluating the function at 2, we get it of the form $\frac{0}{0}$,
Hence,
$$\begin{aligned}& \lim _{x \rightarrow 2} \frac{x^3-2 x^2}{x^2-5 x+6}=\lim _{x\rightarrow 2} \frac{x^2(x-2)}{(x-2)(x-3)} \\& =\lim _{x \rightarrow 2} \frac{x^2}{(x-3)}=\frac{(2)^2}{2-3}=\frac{4}{-1}=-4 .\end{aligned}$$
8. Find the derivative of $4 \sqrt{x}-2$ .
Ans. Let, $f(x)=4 \sqrt{x}-2$
$$\begin{aligned}& f^{\prime}(x)=\frac{d}{d x}(4 \sqrt{x}-2)=\frac{d}{d x}(4 \sqrt{x})-\frac{d}{d x}(2) \\& =4 \frac{d}{d x}\left(x^{\frac{1}{2}}\right)-0=4\left(\frac{1}{2} x^{\frac{1}{2}}\right) \\& =\left(2 x^{-\frac{1}{2}}\right)=\frac{2}{\sqrt{x}}\end{aligned}$$
9. Find the value of $lim_{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x}$ .
Ans. Evaluating the function at 2, we get it of the form $\frac{0}{0}$,
Hence,
$$\begin{aligned}& \lim _{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x}=\lim _{x\rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^2} \\& =\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0} \text { which is not }\end{aligned}$$
Defined.
10. Evaluate: $\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-1}{x}$
Ans. Put $y=1+x$, so that $y \rightarrow 1$ as $x \rightarrow 0$.
Then, $\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\lim _{y \rightarrow 1} \frac{\sqrt{y}-1}{y-1}$
$$\begin{aligned}& =\lim _{y \rightarrow 1} \frac{y^{\frac{1}{2}}-1^{\frac{1}{2}}}{y-1} \\& =\frac{1}{2}(1)^{\frac{1}{2}-1}=\frac{1}{2}\end{aligned}$$
11. Find $\lim_{x\rightarrow 1}f(x)$ where $f(x)= \begin{cases}x^2-1, & x \leq 1 \\-x^2-1, & x>1\end{cases}$
Ans. The given functions is
$$\begin{aligned}& f(x)= \begin{cases}x^2-1, & x \leq 1 \\-x^2-1, & x>1\end{cases} \\& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left[x^2-1\right]=1^2-1=1-1=0 \\& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left[-x^2-1\right]=-1^2-1=-1-1=-2\end{aligned}$$
It is observed that $\lim _{x \rightarrow \Gamma^{-}} f(x) \neq \lim _{x \rightarrow1^{-}} f(x)$.
Hence, $\lim _{x \rightarrow 1} f(x)$ does not exist.
12. Evaluate: $\lim _{x \rightarrow \dfrac{\pi}{6}} \dfrac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}$
Ans. $$ \begin{aligned} & \text { } \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1} \\ & =\lim _{x \rightarrow \frac{\pi}{6}} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)}=\lim _{x \rightarrow \frac{\pi}{6}} \frac{\sin x+1}{\sin x-1} \\ & =\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=-3 \end{aligned} $$
PDF Summary - Class 11 Maths Limits and Derivatives (Chapter 12)
Limits:
Consider a function $f(x)={{x}^{2}}$. Plotting it gives
Here, the value of x approaches 0 as the value of function f(x) moves to 0.
In general as \[x \to a,\text{ }f\left( x \right)\to l\] , then \[l\] is called limit of the function \[f\left( x \right)\] . This is written symbolically as \[\displaystyle \lim_{x \to a}f(x)=l\].
Irrespective of the limits, the function should assume at a given point \[x=a\].
There are two ways in which $x$ can approach a number. It can either be from left or from right. This means that all the values of $x$ near $a$ could be less than $a$ or could be greater than $a$.
Right hand limit - Value of \[f\left( x \right)\] which is dictated by values of \[f\left( x \right)\] when $x$ tends to from the right. It is written as $\displaystyle \lim_{x \to {{a}^{+}}}f(x)$.
Left hand limit - Value of \[f\left( x \right)\] which is dictated by values of \[f\left( x \right)\] when $x$ tends to from the left. It is written as $\displaystyle \lim_{x \to {{a}^{-}}}f(x)$.
Here, the right and left hand limits are different. So, the limit of \[f\left( x \right)\] as \[x\] tends to zero does not exist (even though the function is defined at \[0\]).
If the right and left hand limits coincide then the common value is the limit and denoted by $\displaystyle \lim_{x \to a}f(x)$.
Algebra of limits:
Theorem 1:
Let $f$ and $g$ be two functions such that both \[\displaystyle \lim_{x \to a}f(x)\text{ and }\displaystyle \lim_{x \to a}g(x)\] exist, then
Limit of sum of two functions is sum of the limits of the functions, i.e.
\[\displaystyle \lim_{x \to a}\left[ f(x)+g(x) \right]=\displaystyle \lim_{x \to a}f(x)+\displaystyle \lim_{x \to a}g(x)\] .
Limit of difference of two functions is difference of the limits of the functions, i.e.
\[\displaystyle \lim_{x \to a}\left[ f(x)-g(x) \right]=\displaystyle \lim_{x \to a}f(x)-\displaystyle \lim_{x \to a}g(x)\]
Limit of product of two functions is product of the limits of the functions, i.e.,
\[\displaystyle \lim_{x \to a}\left[ f(x).g(x) \right]=\displaystyle \lim_{x \to a}f(x).\displaystyle \lim_{x \to a}g(x)\]
Limit of quotient of two functions is quotient of the limits of the functions (whenever the denominator is non zero), i.e.,
\[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle \lim_{x \to a}f(x)}{\displaystyle \lim_{x \to a}g(x)}\]
In particular as a special case of (iii), when g is the constant function such that \[g\left( x \right)=\lambda \] , for some real number \[\lambda \] , we have
\[\displaystyle \lim_{x \to a}\left[ \left( \lambda .f \right)(x) \right]=\lambda .\displaystyle \lim_{x \to a}f(x)\]
Limits of polynomials and rational functions:
A function \[f\] is said to be a polynomial function if \[f\left( x \right)\] is zero function or if \[f\left(x\right)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}\] where ${{a}_{i}}S$ is are real numbers such that ${{a}_{n}}\ne 0$ for some natural number $n$.
We know that \[\displaystyle \lim_{x \to a}x=a\]
\[\displaystyle \lim_{x \to a}{{x}^{2}}=\displaystyle \lim_{x \to a}\left( x.x \right)=\displaystyle \lim_{x \to a}x.\displaystyle \lim_{x \to a}x=a.a={{a}^{2}}\]
Hence,
\[\displaystyle \lim_{x \to a}{{x}^{n}}={{a}^{n}}\]
Let \[f\left( x \right)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}\] be a polynomial function
\[\displaystyle \lim_{x \to a}f\left( x \right)=\displaystyle \lim_{x \to a}\left[ {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}} \right]\]
$=\displaystyle \lim_{x \to a}{{a}_{0}}+\displaystyle \lim_{x \to a}{{a}_{1}}x+\displaystyle \lim_{x \to a}{{a}_{2}}{{x}^{2}}+...+\displaystyle \lim_{x \to a}{{a}_{n}}{{x}^{n}}$
$={{a}_{0}}+{{a}_{1}}\displaystyle \lim_{x \to a}x+{{a}_{2}}\displaystyle \lim_{x \to a}{{x}^{2}}+...+{{a}_{n}}\displaystyle \lim_{x \to a}{{x}^{n}}$
\[={{a}_{0}}+{{a}_{1}}a+{{a}_{2}}{{a}^{2}}+...+{{a}_{n}}{{a}^{n}}\]
$=f(a)$
A function $f$ is said to be a rational function, if $f(x)=\dfrac{g(x)}{h(x)}$ where \[g\left( x \right)\text{ and }h\left( x \right)\] are polynomials such that \[h\left( x \right)\ne 0\] .
Then
$\displaystyle \lim_{x \to a}f(x)=\displaystyle \lim_{x \to a}\dfrac{g(x)}{h(x)}=\dfrac{\displaystyle \lim_{x \to a}g(x)}{\displaystyle \lim_{x \to a}h(x)}=\dfrac{g(a)}{h(a)}$
However, if $h(a)=0$ , there are two scenarios –
when \[g\left( a \right)\ne 0\]
limit does not exist
When $g(a)=0$ .
\[g\left( x \right)={{\left( xa \right)}^{k}}{{g}_{1}}\left( x \right)\] , where $k$ is the maximum of powers of $\left( x-a \right)$ in $g(x)$ .
Similarly, \[h\left( x \right)={{\left( xa \right)}^{l}}{{h}_{1}}\left( x \right)\] as $h(a)=0$ . Now, if \[k\ge l\] , we have
$\displaystyle \lim_{x \to a}f(x)=\dfrac{\displaystyle \lim_{x \to a}g(x)}{\displaystyle \lim_{x \to a}h(x)}=\dfrac{\displaystyle \lim_{x \to a}{{\left( xa \right)}^{k}}{{g}_{1}}\left( x \right)}{\displaystyle \lim_{x \to a}{{\left( xa \right)}^{l}}{{h}_{1}}\left( x \right)}$
\[=\dfrac{\displaystyle \lim_{x \to a}{{\left( xa \right)}^{\left( k-l \right)}}{{g}_{1}}\left( x \right)}{\displaystyle \lim_{x \to a}{{h}_{1}}\left( x \right)}=\dfrac{0.{{g}_{1}}(a)}{{{h}_{1}}(a)}=0\]
If \[k < l\] , the limit is not defined.
Theorem 2:
For any positive integer $n$, $\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$.
The proof is shown below.
Dividing $\left( {{x}^{n}}-{{a}^{n}} \right)$ by $\left( x-a \right)$,
$\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=\displaystyle \lim_{x \to a}\left( {{x}^{n-1}}+{{x}^{n-2}}a+{{x}^{n-3}}{{a}^{2}}+...+x{{a}^{n-2}}+{{a}^{n-1}} \right)$
$={{a}^{n-1}}+a\,{{a}^{n-2}}+...+{{a}^{n-2}}(a)+{{a}^{n-1}}$
$={{a}^{n-1}}+{{a}^{n-1}}+...+{{a}^{n-1}}+{{a}^{n-1}}\,\left( n\text{ terms} \right)$
$=n{{a}^{n-1}}$
The expression in the above theorem for the limit is true even if $n$ is any rational number and $a$ is positive.
Limits of Trigonometric Functions:
Theorem 3:
Let $f$ and $g$ be two real valued functions with the same domain such that \[f\left( x \right)\le g\left( x \right)\] for all $x$ in the domain of definition,
For some $a$ , if both $\displaystyle \lim_{x \to a}f(x)\text{ and }\displaystyle \lim_{x \to a}g(x)$ exist, then $\displaystyle \lim_{x \to a}f(x)\le \displaystyle \lim_{x \to a}g(x)$
Theorem 4 (Sandwich Theorem):
Let $f,g$ and $h$ be real functions such that \[f\left( x \right)\le g\left( x \right)\le h\left( x \right)\] for all $x$ in the common domain of definition.
For some real number $a,\text{ if }\displaystyle \lim_{x \to a}f(x)=l=\displaystyle \lim_{x \to a}g(x)$ , then $\displaystyle \lim_{x \to a}g(x)=l$ .
To prove:
$\cos x < \dfrac{\sin x}{x} < 1$ for $0 < \left| x \right| < \dfrac{\pi }{2}$
Proof: Use known facts that \[\sin \left( \text{ }x \right)=\sin x\] and \[\cos \left( \text{ }x \right)=\cos x\]. Hence, it is sufficient to prove the inequality for $0 < x < \dfrac{\pi }{2}$
From the figure, it is noted that O is the centre of the unit circle such that the angle \[AOC\] is $x$ radians and $0 < x < \dfrac{\pi }{2}$ .
Two perpendiculars to \[OA\] are the line segments \[BA\] and \[CD\].
Now join \[AC\] and then,
\[\text{Area of }\Delta OAC < \text{Area of sector }OAC < \text{Area of }\Delta OAB\]
i.e., $\dfrac{1}{2}OA.CD < \dfrac{x}{2\pi }.\pi .{{\left( OA \right)}^{2}} < \dfrac{1}{2}OA.AB$
i.e., $CD < x.OA < AB$
From $\Delta OCD$,
$\sin x=\dfrac{CD}{OA}$ (since $OC=OA$) and hence $CD=OA\sin x$.
Also $\tan x=\dfrac{AB}{OA}$ and hence $AB=OA\tan x$. Thus
$OA\sin x < OA.x < OA\tan x$
Since $OA$ is positive,
$\sin x < x < \tan x$
Since $0 < x < \dfrac{\pi }{2}$, $\sin x$ is positive and thus by dividing throughout by $\sin x$,
$1 < \dfrac{x}{\sin x} < \dfrac{1}{\cos x}$.
Taking reciprocals throughout,
$\sin x < x < \tan x$ .
Since $0 < x < \dfrac{\pi }{2}$, $\sin x$ is positive and thus by dividing throughout by $\sin x$,
$1 < \dfrac{x}{\sin x} < \dfrac{1}{\cos x}$.
Taking reciprocals throughout,
$\cos x < \dfrac{\sin x}{x} < 1$
Hence, Proved.
The following are two important limits
(i). $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$
(ii). $\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{x}=0$
The proof is given as below,
The function $\dfrac{\sin x}{x}$ is sandwiched between the function $\cos x$ and the constant function which takes value $1$.
Since $\displaystyle \lim_{x \to 0}\cos x=1$ and $1-\cos x=2{{\sin }^{2}}\left( \dfrac{x}{2} \right)$,
$\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{x}=\displaystyle \lim_{x \to 0}\dfrac{2{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{x}=\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \dfrac{x}{2} \right)}{\dfrac{x}{2}}.\sin \left( \dfrac{x}{2} \right)$
$=\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{\dfrac{x}{2}}.\displaystyle \lim_{x \to 0}\sin \left( \dfrac{x}{2} \right)=1.0=0$
Using the fact that \[x \to 0\] is equivalent to $\dfrac{x}{2}\to 0$. This may be justified by putting $y=\dfrac{x}{2}$.
Derivatives:
Derivative of a function at a given point in its domain of definition.
Definition 1 - Suppose $f$ is a real valued function and $a$ is a point in its domain of definition. The derivative of $f$ at a is defined by $\displaystyle \lim_{h\to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$, provided this limit exists. ${f}'\left( a \right)$ is used to denoted the derivative of $f\left( x \right)$ at a.
Definition 2 - Suppose $f$ is a real valued function, the function defined by $\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ wherever limit exists is defined to be derivative of $f$ at $x$ denoted by ${f}'\left( x \right)$. This definition of derivative is also called the first principle of derivative.
Thus ${f}'\left( x \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$.
The derivative of function f(x) with respect to x can be denoted in two ways:${f}'\left( x \right)$ is denoted by $\dfrac{d}{dx}\left( f\left( x \right) \right)$ or if $y=f\left( x \right)$, it is denoted by $\dfrac{dy}{dx}$.
Another notation is $D\left( f\left( x \right) \right)$.
Further, derivative of f at $x=a$ is also denoted by \[{{\left. \dfrac{d}{dx}f\left( x \right) \right|}_{a}}\text{ or }{{\left. \dfrac{df}{dx} \right|}_{a}}\text{ or even }{{\left( \dfrac{df}{dx} \right)}_{x=a}}\].
Theorem 5:
Let $f$ and $g$ be two functions such that their derivatives are defined in a common domain. Then
Derivative of sum of two functions is sum of the derivatives of the functions.
\[\dfrac{d}{dx}\left[ f\left( x \right)+g\left( x \right) \right]=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)\]
Derivative of difference of two functions is difference of the derivatives of the functions.
\[\dfrac{d}{dx}\left[ f\left( x \right)-g\left( x \right) \right]=\dfrac{d}{dx}f\left( x \right)-\dfrac{d}{dx}g\left( x \right)\]
Derivative of product of two functions is given by following product rule.
\[\dfrac{d}{dx}\left[ f\left( x \right).g\left( x \right) \right]=\dfrac{d}{dx}f\left( x \right).g(x)+f(x).\dfrac{d}{dx}g\left( x \right)\]
Derivative of quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero).
\[\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{\dfrac{d}{dx}f(x).g(x)-f(x)\dfrac{d}{dx}g(x)}{{{\left( g(x) \right)}^{2}}}\]
Let $u=f(x)$ and $v=g(x)$ .
Product Rule:
${{\left( uv \right)}^{\prime }}={u}'v+u{v}'$ .
Also referred as Leibnitz rule for differentiating product of functions
Quotient Rule:
${{\left( \dfrac{u}{v} \right)}^{\prime }}=\dfrac{{u}'v-u{v}'}{{{v}^{2}}}$
Derivative of the function $f(x)=x$ is the constant.
Theorem 6:
Derivative of \[f(x)={{x}^{n}}\] is \[n{{x}^{n-1}}\] for any positive integer \[n\].
Proof
By definition of the derivative function, we have
${f}'(x)=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\displaystyle \lim_{h\to 0}\dfrac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$ .
Binomial theorem tells that ${{\left( x+h \right)}^{n}}=\left( {}^{n}{{C}_{0}} \right){{x}^{n}}+\left( {}^{n}{{C}_{1}} \right){{x}^{n-1}}h+...+\left( {}^{n}{{C}_{n}} \right){{h}^{n}}$ and ${{\left( x+h \right)}^{n}}-{{x}^{n}}=h\left( n{{x}^{n-1}}+...+{{h}^{n-1}} \right)$. Thus
$\dfrac{df(x)}{dx}=\displaystyle \lim_{h\to 0}\dfrac{{{\left( x+h \right)}^{n}}-{{x}^{n}}}{h}$
$=\displaystyle \lim_{h\to 0}\dfrac{h\left( n{{x}^{n-1}}+...+{{h}^{n-1}} \right)}{h}$
$=\displaystyle \lim_{h\to 0}\left( n{{x}^{n-1}}+...+{{h}^{n-1}} \right)=n{{x}^{n-1}}$
This can be proved as below alternatively
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=\dfrac{d}{dx}\left( x.{{x}^{n-1}} \right)\]
\[=\dfrac{d}{dx}(x).\left( {{x}^{n-1}} \right)+x.\dfrac{d}{dx}\left( {{x}^{n-1}} \right)\] (By product rule)
$=1.{{x}^{n-1}}+x.\left( (n-1){{x}^{n-2}} \right)$ (By induction hypothesis)
$={{x}^{n-1}}+(n-1){{x}^{n-1}}=n{{x}^{n-1}}$
Theorem 7:
Let $f(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+....+{{a}_{1}}x+{{a}_{0}}$ be a polynomial function, where ${{a}_{i}}$ s are all real numbers and ${{a}_{n}}\ne 0$. Then, the derivative function is given by
$\dfrac{df(x)}{dx}=n{{a}_{n}}{{x}^{n-1}}+(n-1){{a}_{n-1}}{{x}^{n-2}}+....+2{{a}_{2}}x+{{a}_{1}}$
Quick Reference:
For functions $f$ and $g$ the following holds:
\[\displaystyle \lim_{x \to a}\left[ f(x)\pm g(x) \right]=\displaystyle \lim_{x \to a}f(x)\pm \displaystyle \lim_{x \to a}g(x)\]
\[\displaystyle \lim_{x \to a}\left[ f(x).g(x) \right]=\displaystyle \lim_{x \to a}f(x).\displaystyle \lim_{x \to a}g(x)\]
\[\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)}=\dfrac{\displaystyle \lim_{x \to a}f(x)}{\displaystyle \lim_{x \to a}g(x)}\]
Following are some of the standard limits
$\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$
$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1,\displaystyle \lim_{x \to a}\dfrac{\sin \left( x-a \right)}{x-a}=1$
$\displaystyle \lim_{x \to 0}\dfrac{1-\cos x}{x}=0$
$\displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}=1,\displaystyle \lim_{x \to a}\dfrac{\tan \left( x-a \right)}{x-a}=1$
$\displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{-1}}x}{x}=1,\displaystyle \lim_{x \to 0}\dfrac{{{\tan }^{-1}}x}{x}=1$
$\displaystyle \lim_{x \to 0}\dfrac{{{a}^{x}}-1}{x}={{\log }_{e}}a,a > 0,a\ne 1$
Derivatives
The derivative of a function $f$ at $a$ is defined by
${f}'\left( a \right)=\displaystyle \lim_{h\to 0}\dfrac{f\left( a+h \right)-f\left( a \right)}{h}$
Derivative of a function $f$ at any point $x$ is defined by
${f}'\left( x \right)=\dfrac{df(x)}{dx}=\displaystyle \lim_{h\to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
For functions $u$ and $v$ the following holds:
${{\left( u\pm v \right)}^{\prime }}={u}'\pm {v}'$
${{\left( uv \right)}^{\prime }}={u}'v+u{v}'\Rightarrow \dfrac{d}{dx}\left( uv \right)=u.\dfrac{dv}{dx}+v.\dfrac{du}{dx}$
${{\left( \dfrac{u}{v} \right)}^{\prime }}=\dfrac{{u}'v-u{v}'}{{{v}^{2}}}\Rightarrow \dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v.\dfrac{du}{dx}-u.\dfrac{dv}{dx}}{{{v}^{2}}}$
Following are some of the standard derivatives
\[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
\[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]
\[\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x\]
\[\dfrac{d}{dx}\left( \cot x \right)=-\text{cose}{{\text{c}}^{2}}x\]
\[\dfrac{d}{dx}\left( \sec x \right)=\sec x.\tan x\]
\[\dfrac{d}{dx}\left( \cos \text{ec }x \right)=-\cos \text{ec }x.\cot x\]
Limits and Derivatives Class 11 Notes – Chapter Overview
Class 11 Revision Notes Limits and Derivatives PDF
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Class 11 Maths Revision Notes Chapter 12
Limits - In Calculus, we study how the value of a function changes with a change in the points in the domain. The limit of a function is stated as x -> p, f(x) -> 1, where 1 is the limit of the function f(x). Its notation is: x pf(x) = l. Some key points about the limit of a function are:
X could approach a number from either the left or right, i.e., all the values that x could assume near p could be either less than or greater than p.
Limits are of two types:
a. Right-hand limit - When x tends to p from the right, then that f(x) value is the right-hand limit.
b. Left-hand limit - When x tends to p from the left, then that value of f(x) is the left-hand limit.
Standard Limits - Listed Below Are Some Standard Limits
\[\lim_{x \to p}\] (xn - pn)/x - p = npn-1
\[\lim_{x \to 0}\] (sin x)/x = 1
\[\lim_{x \to p}\] (sin (x - p))/(x - p) = 1
\[\lim_{x \to 0}\] (tan x)/x = 1
\[\lim_{x \to p}\] (tan (x - p))/(x - p) = 1
\[\lim_{x \to 0}\] (sin-1 x)/x = 1
\[\lim_{x \to 0}\] (tan-1 x)/x = 1
\[\lim_{x \to 0}\] (px - 1)/x = logep, p > 0 and p 1
Algebra of Limits - If f and z are two functions and limits for both exists (\[\lim_{x \to p}\]f(x), \[\lim_{x \to p}\]z(x)), then:
The limit of the sum of two functions is equal to the individual sum of limits of the two functions.
\[\lim_{x \to p}\] [f(x) + z(x)] = \[\lim_{x \to p}\]f(x) + \[\lim_{x \to p}\]z(x)
The limit of the difference of two functions is equal to the difference in limits of the two functions.
\[\lim_{x \to p}\] [f(x) - z(x)] = \[\lim_{x \to p}\]f(x) - \[\lim_{x \to p}\]z(x)
The limit of the multiplication of two functions is equal to the product of the limits of the two functions.
\[\lim_{x \to p}\] [f(x) * z(x)] = \[\lim_{x \to p}\]f(x) * \[\lim_{x \to p}\]z(x)
If z is c constant function and z(x) = λ then:
\[\lim_{x \to p}\] [(λ * f)(x)] = λ * \[\lim_{x \to p}\]f(x)
The limit of the division of two functions is equal to the division of limits of the two functions, provided the denominator is non-zero.
\[\lim_{x \to p}\][f(x)/z(x)] = \[\lim_{x \to p}\]f(x)/\[\lim_{x \to p}\]z(x)
Derivatives - Derivative of a function y = f(x) is found by using the formulas change in y/change in x. Lets us say x changes from x to dx, then y changes from y to f(x) to f(x + dx) so:
Derivative = change in y/change in x = dy/dx = f(x + dx) - f(x)/dx
Derivative of a function at a point p is given by:
f’(p) = \[\lim_{h \to 0}\]f(p + h)/h
Some Standard Derivatives - Mentioned Below Are Some Standard Derivatives
d(xi)/dx = ixi-1
d(sin x)/dx = cos x
d(cos x)/dx = -sin x
d(tan x)/dx = sec2x
d(cot x)/dx = -coses2x
d(sec x)/dx = secx . tanx
d(cosex x)/dx = -cosec x. cot x
What are the Benefits of Referring to Vedantu’s Revision Notes for Class 11 Chapter 12 - Limits and Derivatives
1. Quick Summaries: Get brief, easy-to-understand overviews of main concepts.
2. Simplified Topics: Complex subjects made simpler for better comprehension.
3. Last-Minute Prep: Efficient tool for swift exam preparation.
4. Enhanced Retention: Aid in remembering crucial information effectively.
5. Exam Support: Key points and tips for effective exam readiness.
6. Time-Saving: Consolidated information saves study time.
7. Priority Focus: Emphasis on important topics and questions.
8. Real-world Examples: Practical applications for better understanding.
9. Confidence Boost: Build confidence for better performance in exams.
Conclusion
For an enhanced comprehension of this subject, NCERT - Class 11 Chapter 12, “Limits and Derivatives” thoughtfully prepared by experienced educators at Vedantu is your invaluable companion. These notes break down the complexities of “Limits and Derivative” into easily digestible sections, helping you grasp new concepts, master formulas, and navigate through questions effortlessly quickly in the last minute as well. By immersing yourself in these notes, you not only prepare for your studies more efficiently but also develop a profound understanding of the subject matter.
Important Study Material Links for Chapter 12: Limits and Derivatives
NCERT Class 11 Maths Solutions Chapter-wise Links - Download the FREE PDF
Important Related Links for CBSE Class 11 Maths
FAQs on Limits and Derivatives Class 11 Notes CBSE Maths Chapter 12 (Free PDF Download)
1. What are some of the real-world applications of the concept of derivatives?
Some examples from real-world where the theory of derivatives is useful are:
Those who maintain a reservoir need to know when it is going to overflow. This is done by measuring the depth of the water in several instances of time.
By knowing the height of the rocket at different instances of time, rocket scientists can compute the exact velocity with which a satellite should be shot out from a rocket.
Financial analysts can predict the changes in the value of a particular stock in the stock market by knowing its present value.
2. What is the sandwich theorem of limits?
The sandwich theorem of limits states that if f, g, and z are real functions where f(x) <= g(x) <= z(x) for all x in the common domain then for a real number p if
3. What is the use of revision notes of Chapter 12 for Class 11 Mathematics?
Students think that notes are only beneficial for subjects like Chemistry or Biology. However, this is not true. For a subject like Mathematics, you need to understand the fundamentals of a given chapter and then proceed on doing the exercises. Notes of Chapter 12 of Class 11 Mathematics help you put the explanations, theorems, and formulas all in one place. This helps you in understanding and retaining the concepts well. Class 11 Mathematics revision notes for Chapter 12 by Vedantu are available on the page Maths Revision Notes for Class 11 at free of cost. You can also download the Vedantu app to access study material for free.
4. Is Chapter 12 of Class 11 Maths easy?
Class 11 Mathematics may seem challenging for some, but it is not extremely difficult either. It requires time, effort, and understanding on a student's part. Two main strategies of doing well in Chapter 12 of Class 11 Maths are understanding the concepts well and getting a lot of practice. Make efforts to understand the concepts well and dedicate a good amount of your time to practice the NCERT questions, sample papers, and previous years’ questions. This would ease the apparent difficulty of the subject.
5. How to ace Chapter 12 “Limits and Derivatives”?
Calculus studies rates of change and continuity of quantities. “Limits and Derivatives” forms about half of calculus. To ace this chapter:
The most important thing is to pay attention in class and understand the concept well. Refer all your doubts to your teachers at the earliest. You can also watch videos for enhanced understanding.
The more you practice this chapter, the more proficient you will be. Practising sample papers and question papers is a must.
6. What is the correct approach to study Chapter 12 of Class 11 Mathematics?
How you perform in a subject largely depends on the approach you take whilst studying the subject. The same goes for maths. To study Chapter 12, follow these strategies -
Have the correct mindset while studying. Don’t be intimated. If something seems difficult, break it down and study in parts.
Work on understanding the chapter's concepts and applications.
Clarify doubts.
Practice a ton.
Refer to extra resources like revision notes, NCERT Solutions, important theorems, and conceptual videos to advance your grasp of the topics.
7. What is the objective of studying limits and derivatives?
Understanding the real-time applications of concepts in Mathematics helps in understanding the chapter well and in turn performing well in the exams. Limits and Derivatives have important applications, not only in advanced Mathematics but also in other streams like Physics and Engineering. They can be used to calculate electric and magnetic fields, finding the rate of change of a quantity, checking temperature variations, etc. You can refer to Vedantu's Revision Notes Chapter 12 to gain a better understanding of the chapter.
