Limits and Derivatives Class 11 Notes CBSE Maths Chapter 12 (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

1.  \[\lim_{x \to p}\] (xn - pn)/x - p = npn-1

2.  \[\lim_{x \to 0}\] (sin x)/x = 1

3. \[\lim_{x \to p}\] (sin (x - p))/(x - p) = 1

4. \[\lim_{x \to 0}\] (tan x)/x = 1

5. \[\lim_{x \to p}\] (tan (x - p))/(x - p) = 1

6. \[\lim_{x \to 0}\] (sin-1 x)/x = 1

7. \[\lim_{x \to 0}\] (tan-1 x)/x = 1

8. \[\lim_{x \to 0}\] (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

1. d(xi)/dx = ixi-1

2. d(sin x)/dx = cos x

3. d(cos x)/dx = -sin x

4. d(tan x)/dx = sec2x

5. d(cot x)/dx = -coses2x

6. d(sec x)/dx = secx . tanx

7. 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.

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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

• Limits and Derivatives Important Questions

• Limits and Derivatives Important Formulas

• Limits and Derivatives Revision Notes

• Limits and Derivatives NCERT Exemplar Solutions

• Limits RD Sharma Solutions

• Derivatives RD Sharma Solutions

• Limits RS Aggarwal Solutions

NCERT Class 11 Maths Solutions Chapter-wise Links - Download the FREE PDF

Important Related Links for CBSE Class 11 Maths