This MCQ module is based on: Limits of Trigonometric Functions
TOPIC 38 OF 45
Limits of Trigonometric Functions
🎓 Class 11
Mathematics
CBSE
Theory
Ch 12 — Limits and Derivatives
⏱ ~30 min
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This mathematics assessment will be based on: Limits of Trigonometric Functions
Targeting Class 11 level in Calculus, with Advanced difficulty.
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12.4 Limits of Trigonometric Functions
Trigonometric functions such as \(\sin x\) and \(\cos x\) are continuous everywhere on \(\mathbb R\). This means that for any real number \(a\): \[\lim_{x\to a}\sin x=\sin a,\qquad \lim_{x\to a}\cos x=\cos a.\] However, many limits involving trig functions produce the indeterminate form \(\tfrac{0}{0}\) — notably \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x}\). To handle these, we prove one foundational result known as the Sandwich Theorem?.
Sandwich (Squeeze) Theorem
Let \(f\), \(g\), \(h\) be three functions such that \(f(x)\le g(x)\le h(x)\) for all \(x\) in a deleted neighbourhood of \(a\). If \(\displaystyle \lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L\), then \(\displaystyle \lim_{x\to a}g(x)=L\).
A Fundamental Limit: \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1\)
We prove this using the sandwich theorem and a geometric inequality.
Fig 12.3: Unit circle, with \(\sin x\) (vertical red segment), \(x\) (arc length), and \(\tan x\) (green tangent segment).
For \(0 Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin 4x}{\sin 2x}\). Evaluate \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^2}\). Evaluate \(\displaystyle \lim_{x\to 0}\frac{\tan 3x}{\sin 5x}\). Confirm \(\lim_{x\to 0}\frac{\sin x}{x}=1\) numerically.
Standard Trigonometric Limit
\[\lim_{x\to 0}\frac{\sin x}{x}=1, \qquad \lim_{x\to 0}\frac{1-\cos x}{x}=0, \qquad \lim_{x\to 0}\frac{\tan x}{x}=1.\]
Here \(x\) is measured in radians.
Example 9
\(\dfrac{\sin 4x}{\sin 2x}=\dfrac{\sin 4x}{4x}\cdot\dfrac{2x}{\sin 2x}\cdot\dfrac{4x}{2x}=\dfrac{\sin 4x}{4x}\cdot\dfrac{2x}{\sin 2x}\cdot 2\).
As \(x\to 0\): \(1\cdot 1\cdot 2 = 2\).Example 10
Use \(1-\cos x = 2\sin^2(x/2)\): \(\dfrac{1-\cos x}{x^2}=\dfrac{2\sin^2(x/2)}{x^2}=\dfrac{1}{2}\left(\dfrac{\sin(x/2)}{x/2}\right)^2\).
As \(x\to 0\), \(\dfrac{\sin(x/2)}{x/2}\to 1\), so limit \(=\dfrac{1}{2}\).Example 11
\(\dfrac{\tan 3x}{\sin 5x}=\dfrac{\tan 3x}{3x}\cdot\dfrac{5x}{\sin 5x}\cdot\dfrac{3}{5}\).
Each of \(\tfrac{\tan 3x}{3x}\to 1\), \(\tfrac{5x}{\sin 5x}\to 1\). Limit \(=\dfrac{3}{5}\).Example 12 (numerical check)
x (rad) 0.5 0.1 0.01 0.001 -0.001 -0.01 sin x / x 0.9589 0.99833 0.99998 0.9999998 0.9999998 0.99998 Exercise 12.1 (selected trig items)
Q1. Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin ax}{bx}\), \(b\ne 0\).
\(\dfrac{\sin ax}{bx}=\dfrac{\sin ax}{ax}\cdot\dfrac{a}{b}\to \dfrac{a}{b}\).Q2. Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin ax}{\sin bx}\), \(a,b\ne 0\).
\(=\dfrac{\sin ax}{ax}\cdot\dfrac{bx}{\sin bx}\cdot\dfrac{a}{b}\to \dfrac{a}{b}\).Q3. Evaluate \(\displaystyle \lim_{x\to 0}\frac{1-\cos 4x}{1-\cos 6x}\).
Using Example 10 idea: \(\dfrac{1-\cos 4x}{x^2}\to 8\) and \(\dfrac{1-\cos 6x}{x^2}\to 18\).
Ratio \(\to \dfrac{8}{18}=\dfrac{4}{9}\).Q4. Evaluate \(\displaystyle \lim_{x\to 0}\frac{\cos x}{\pi - x}\).
Both numerator and denominator are continuous and \(\pi-0=\pi\ne 0\). Limit \(=\dfrac{1}{\pi}\).Q5. Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(\pi-x)}{\pi(\pi-x)}\).
\(\sin(\pi-x)=\sin x\). Limit \(=\displaystyle\lim_{x\to 0}\dfrac{\sin x}{\pi(\pi-x)}=\dfrac{0}{\pi\cdot\pi}=0\).Q6. Evaluate \(\displaystyle \lim_{x\to 0}\frac{\sin(x-a)}{\cos(x-a)}\) carefully.
This is \(\tan(x-a)\), continuous at \(x=0\) (provided \(\cos a\ne 0\)). Limit \(=\tan(-a)=-\tan a\).Q7. Let \(f(x)=\begin{cases}\dfrac{\sin x}{x},&x\ne 0\\ 1,&x=0\end{cases}\). Is \(\lim_{x\to 0}f(x)=f(0)\)?
Yes: \(\lim_{x\to 0}\dfrac{\sin x}{x}=1=f(0)\). The piecewise definition has been chosen precisely to make \(f\) continuous at 0.
Activity: The Sandwich on a CalculatorL3 ApplyMaterials: Scientific calculator (set to RADIAN mode!), table-sheet.
Predict: For very small \(x\) (in radians), what is the approximate value of \(\dfrac{\sin x}{x}\)?
In radian mode the limit is 1. In degree mode, \(\lim_{x\to 0}\dfrac{\sin x^\circ}{x}=\dfrac{\pi}{180}\approx 0.01745\). The difference arises because the geometric inequality \(\sin x
Competency-Based Questions
Scenario: A student is evaluating several trigonometric limits for homework. She wants to apply \(\lim_{x\to 0}\tfrac{\sin x}{x}=1\) efficiently.
Q1. \(\displaystyle \lim_{x\to 0}\frac{\sin 7x}{x}\) equals:L3 ApplyAnswer: (b) 7. \(\dfrac{\sin 7x}{x}=7\cdot\dfrac{\sin 7x}{7x}\to 7\cdot 1=7\).Q2. Analyse: why must we multiply and divide by the argument of sine when using the standard limit?L4 AnalyseAnswer: The standard result \(\lim_{u\to 0}\tfrac{\sin u}{u}=1\) requires the ratio to have the same argument in numerator and denominator. By writing \(\dfrac{\sin kx}{x}=k\cdot\dfrac{\sin kx}{kx}\), we create the ratio \(\tfrac{\sin(kx)}{kx}\) whose limit is 1, leaving the numerical factor \(k\).Q3. Evaluate the claim: "\(\displaystyle \lim_{x\to 0}\dfrac{\sin x}{x^2}\) equals 1 since the standard limit is 1."L5 EvaluateAnswer: The claim is false. Write \(\dfrac{\sin x}{x^2}=\dfrac{\sin x}{x}\cdot\dfrac{1}{x}\). The first factor tends to 1 but the second factor is \(\tfrac{1}{x}\), which tends to \(+\infty\) as \(x\to 0^+\) and \(-\infty\) as \(x\to 0^-\). So the limit does not exist. The student confused \(x^2\) in the denominator with \(x\).Q4. Create a limit problem that evaluates to \(\tfrac{5}{3}\) using only the standard trig limits. Provide full solution.L6 CreateSample: \(\displaystyle \lim_{x\to 0}\frac{\sin 5x}{\tan 3x}\). Rewrite: \(=\dfrac{\sin 5x}{5x}\cdot\dfrac{3x}{\tan 3x}\cdot\dfrac{5}{3}\to 1\cdot 1\cdot\dfrac{5}{3}=\dfrac{5}{3}\). Other valid creations: \(\lim\tfrac{\sin 10x}{\sin 6x}\), etc.
Assertion–Reason Questions
Assertion (A): \(\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1\) when \(x\) is in radians.
Reason (R): In the unit circle, for small positive \(x\), \(\cos x \le \frac{\sin x}{x}\le 1\), and \(\cos x\to 1\).Answer: (a). R is exactly the sandwich-theorem derivation that gives A.
Assertion (A): \(\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x}=0\).
Reason (R): \(1-\cos x = 2\sin^2(x/2)\).Answer: (a). Using R: \(\dfrac{1-\cos x}{x}=\dfrac{2\sin^2(x/2)}{x}=\sin(x/2)\cdot\dfrac{\sin(x/2)}{x/2}\to 0\cdot 1=0\).
Assertion (A): \(\displaystyle \lim_{x\to 0}\frac{\tan x}{x}=1\).
Reason (R): \(\tan x=\tfrac{\sin x}{\cos x}\) and \(\cos x\to 1\).Answer: (a). \(\tfrac{\tan x}{x}=\tfrac{\sin x}{x}\cdot\tfrac{1}{\cos x}\to 1\cdot 1 = 1\). R explains A.Frequently Asked Questions
What is limit as x approaches 0 of (1 - cos x)/x?
The limit equals 0. It can be derived from the sin x / x limit using the identity 1 - cos x = 2 sin^2(x/2).What is limit as x approaches 0 of tan x / x?
The limit equals 1. Since tan x = sin x / cos x, this follows from limit sin x / x = 1 and limit cos x = 1 as x approaches 0.Why must x be in radians for these standard limits?
The identity sin x is approximately x for small x (and hence sin x / x is approximately 1) holds only when x is in radians; in degrees the ratio is pi/180.How does the sandwich (squeeze) theorem help?
It proves limit sin x / x = 1 by trapping (sin x)/x between cos x and 1 for small positive x and taking x to 0.What is limit as x approaches 0 of sin(k x)/x?
It equals k, for any constant k. Write sin(k x)/x = k * sin(k x)/(k x) and apply the standard limit.Are trigonometric limits in the JEE syllabus?
Yes. Standard trig limits appear directly in JEE Main and also support continuity and differentiability problems in JEE Advanced.
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