How to evaluate trigonometric limits using Sandwich Theorem with proof and examples
We will discuss a crucial idea utilised in calculus and its limits in this topic. We'll talk about the Sandwich theorem notion. This theorem is very easy to comprehend and has numerous calculus applications. It is particularly useful for applying trigonometric functions. A thorough comprehension of the sandwich theorem will be useful not only for calculus but also for other ideas like the binomial theorem.
History of Carl Friedrich Gauss
In an effort to compute π, the mathematicians Archimedes and Eudoxus utilised the Sandwich theorem for the first time geometrically. Carl Friedrich Gauss then expressed it in more contemporary terms. Johann Carl Friedrich Gauss was a German mathematician and physicist who made a significant impact on a variety of scientific and mathematical subjects.
Carl Friedrich Gauss (30 April 1777 – 23 February 1855)
What is the Sandwich Theorem in Limits?
Let f, g, and h be real functions such that for any x in the shared definition domain, f(x) g(x) h(x) occurs. When \[a\] is a real number, then
\[\begin{array}{l}\mathop {{\rm{\;}}\lim }\limits_{x \to a} f\left( x \right)\;\; = \;\mathop {lim}\limits_{x \to a} h\left( x \right)\; = \;t\\{\rm{then}}\;\mathop {\lim }\limits_{x \to a} g\left( x \right)\; = \;t\end{array}\]
Sandwich Theorem in Limits
Sandwich Theorem Proof
Let's examine the geometric proof for the aforementioned claim using a trigonometric inequality.
Think of the disparity as
\[\cos x\; < \;\sin x\;/\;x\; < 1\]
Sandwich Theorem Proof
We can see the triangles ΔABE, ΔADF, ΔADB, and sector ADB in the above diagram.
\[\begin{array}{l}Area\;\left( {\Delta ABD} \right)\; < \;Area\;\left( {sector\;ADB} \right)\; < \;Area\;\left( {\Delta ADF} \right)\\\; \Rightarrow 1/2.AD.EB\; < \;x/2\pi .\pi .A{D^2}\; < \;1/2.AD.DF\end{array}\]
By eliminating the common terms from all sides,
\[{\rm{EB}} < x.AD < {\rm{ DF}}\]
Since angle A = angle X,
\[EB\; = \;AB\; \sin x\] (from Δ ABE),
and \[\sin A\; = \;EB/AB\].
\[DF\; = \;AD \tan X\] because \[\tan A\; = \;DF/AD\].
However, \[\tan A\; = \;\sin X/\cos x\;\] and \[AB\; = \;AD\].
Consequently,
\[\begin{array}{l}AD.\sin A\; < x.AD < \;AD.\sin A\;/\;\cos A\\ \Rightarrow \; = \;1\; < x\;/\;\sin X < \;1\;/\;\cos X\end{array}\]
By taking reciprocals,
\[\cos x\; < \;\sin x\;/\;x\; < 1\]
Hence proved.
We can simply demonstrate some other trigonometric identities using the above theory, including
\[\begin{array}{l}li{m_{x - > 0}}\sin x/x\; = 1\;\\li{m_{x - > 0}}\;\left( {1\;--\;\cos x} \right)/x\; = 0\end{array}\]
Limitations of Sandwich Theorem
The Sandwich theorem is not applicable if left and right function limits are not equal.
The Sandwich Theorem does not apply if we know the limits of any of two functions other than both extreme functions.
Applications of Sandwich Theorem
To determine the limits of specific trigonometric functions, we can utilise the Sandwich rule.
It can be applied to link certain sequences between other known sequences that also converge to the same place in order to demonstrate the convergence of those sequences.
Interesting Fact
If \[f\left( x \right) \le g\left( x \right)\] when x is close to \[\alpha \] (unless perhaps at an \[\alpha \]), then $\underset{x\to \alpha }{\mathop{\lim }}\,f\left( x \right)~\underset{x\to \alpha }{\mathop{\lim }}\,g\left( x \right)$.
Solved Examples of Limits of Trigonometric Functions Using Sandwich Theorem
1. Evaluate the limit for the following tangent function \[lim{_{\;x - > 0}}\;\tan x\;/\;x\].
Ans: Using the trigonometric identity,
\[\begin{array}{l}\tan x\; = \dfrac{{\;\sin x}}{{\cos x}}\\\therefore \mathop {\lim }\limits_{x \to 0} \;\dfrac{{\sin x}}{{x\cos x}}\; = \;\mathop {\lim }\limits_{x \to 0} \;\dfrac{{\sin x}}{x}\cdot\mathop {\lim }\limits_{x \to 0} \;\dfrac{1}{{\cos x}}\end{array}\]
And from the Sandwich theorem, we know,
\[\begin{array}{l}li{m_{x - > 0}}\sin x/x\; = 1\\\;li{m_{x - > 0}}\;1/\cos x\; =1\end{array}\]
Therefore,
\[1\cdot\,1\; = \;1\]
2. Prove \[\mathop {\lim }\limits_{x \to 0} \sin x\; = \;0\].
Ans. As known,
$-x\leqslant\sin x\leqslant x\, for \, all\,x \geqslant 0$
\[\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 0} \left( { - x} \right)\; = \;0\;\;{\rm{and}}\\\mathop {\lim }\limits_{x \to 0} \left( x \right)\; = \;0\end{array}\]
By Sandwich Theorem,
\[\mathop {\lim }\limits_{x \to 0} \left( {Sinx} \right)\; = \;0\]
3. Evaluate \[\mathop {\lim }\limits_{x \to 0} x\dfrac{{secx - 1}}{x}\]
Ans. As we know,
\[\sec x\; = \dfrac{{\;1}}{x}\]
So,
\[\begin{array}{l}\mathop {\lim }\limits_{x \to 0} x\dfrac{{secx - 1}}{x} = \mathop {\lim }\limits_{x \to 0} x\dfrac{{\dfrac{1}{{\cos x}} - 1}}{x}\\\mathop {\lim }\limits_{x \to 0} x\dfrac{{1\; - \cos x}}{{x\cos x}}\\\mathop { \Rightarrow \lim }\limits_{\;\;\;\;\;\;x \to 0} x\left( {\dfrac{1}{{\cos x}}} \right)\left( {\dfrac{{1\; - \cos x}}{x}} \right)\\ \Rightarrow \mathop {\lim }\limits_{x \to 0} x\left( {\dfrac{1}{{\cos x}}} \right)\mathop {\lim }\limits_{x \to 0} x\left( {\dfrac{{1\; - \cos x}}{x}} \right)\\ \Rightarrow 1 \times 0\\ \Rightarrow 0\end{array}\]
Conclusion
In this article, we have thoroughly discussed the Sandwich theorem and its proof. From the discussion above, about the Sandwich theorem we can conclude that the sandwich theorem allows us to determine the limit of a single function by using the limits of the two other functions that it is "sandwiched" between as a starting point. When a function gets complex or intricate, or perhaps when we are unable to identify the limit using other techniques, we apply the Sandwich theorem to determine its limit. The limits of trigonometric functions are most frequently found using this method.
Important Points
This theorem is probably utilised to establish the limit of the function in question.
All the functions considered must be real.
Important Formula From the Theorem
According to the Sandwich theorem,
If \[\begin{array}{l}\mathop {{\rm{\;}}\lim }\limits_{x \to a} f\left( x \right)\;\; = \;\mathop {lim}\limits_{x \to a} h\left( x \right)\; = \;t\\{\rm{then}}\;\mathop {\lim }\limits_{x \to a} g\left( x \right)\; = \;t\end{array}\]
FAQs on Limits of Trigonometric Functions by Sandwich Theorem Explained
1. What is the Sandwich Theorem in limits?
The Sandwich Theorem (also called the Squeeze Theorem) states that if a function f(x) is trapped between two functions g(x) and h(x), and both have the same limit at a point, then f(x) has that same limit.
- If g(x) ≤ f(x) ≤ h(x) near x = a
- And limₓ→a g(x) = limₓ→a h(x) = L
- Then limₓ→a f(x) = L
2. How do you use the Sandwich Theorem to find limits of trigonometric functions?
To use the Sandwich Theorem for trigonometric limits, bound the trigonometric expression between two simpler functions with known limits.
- Step 1: Find inequalities involving sin x or cos x.
- Step 2: Divide or manipulate the inequality properly.
- Step 3: Take limits of the bounding functions.
- Step 4: Conclude the limit of the middle function.
3. Why is the limit of sin x divided by x equal to 1?
The limit limₓ→0 (sin x)/x = 1 because the function is squeezed between cos x and 1, both of which approach 1 as x → 0.
- cos x ≤ (sin x)/x ≤ 1
- limₓ→0 cos x = 1
- limₓ→0 1 = 1
4. What is the limit of x sin(1/x) as x approaches 0 using the Sandwich Theorem?
The limit limₓ→0 x sin(1/x) = 0 using the Sandwich Theorem.
- Since −1 ≤ sin(1/x) ≤ 1
- Multiply by x: −|x| ≤ x sin(1/x) ≤ |x|
- As x → 0, both −|x| and |x| → 0
5. What are the important trigonometric inequalities used in the Sandwich Theorem?
The most important inequality for trigonometric limits is cos x ≤ (sin x)/x ≤ 1 for small x ≠ 0. Other useful inequalities include:
- −1 ≤ sin x ≤ 1
- −1 ≤ cos x ≤ 1
6. Can you give a simple example of a trigonometric limit solved using the Sandwich Theorem?
Yes, a common example is finding limₓ→0 x² cos(1/x).
- Since −1 ≤ cos(1/x) ≤ 1
- Multiply by x²: −x² ≤ x² cos(1/x) ≤ x²
- As x → 0, both −x² and x² → 0
7. When can the Sandwich Theorem be applied in trigonometric limits?
The Sandwich Theorem can be applied when a trigonometric function is bounded between two functions that have the same limit at a point. The conditions are:
- g(x) ≤ f(x) ≤ h(x) near x = a
- limₓ→a g(x) = limₓ→a h(x)
8. What is the difference between the Sandwich Theorem and direct substitution in limits?
Direct substitution works when the function is continuous, while the Sandwich Theorem is used when direct substitution fails or the function oscillates.
- Direct substitution: Used if the function is defined and continuous at a point.
- Sandwich Theorem: Used when expressions like sin(1/x) do not have a direct limit.
9. Why is the Sandwich Theorem important for proving standard trigonometric limits?
The Sandwich Theorem is important because it rigorously proves fundamental results like limₓ→0 (sin x)/x = 1. These standard limits are used to:
- Define derivatives of trigonometric functions
- Develop calculus formulas
- Solve advanced limit problems
10. What are common mistakes when using the Sandwich Theorem for trigonometric limits?
Common mistakes include not verifying the inequality correctly or assuming limits without checking both bounds. Key errors to avoid:
- Forgetting to ensure g(x) ≤ f(x) ≤ h(x)
- Not confirming both outer limits are equal
- Incorrectly multiplying inequalities when x is negative
