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1.3 Q-3

Question Statement

Evaluate the following limits:

i. limh0sin7xx\quad \lim _{h \rightarrow 0} \frac{\sin 7 x}{x}

ii. limx0sinx0x\quad \lim _{x \rightarrow 0} \frac{\sin x^{0}}{x}

iii. limx01cosθsinθ\quad \lim _{x \rightarrow 0} \frac{1-\cos \theta}{\sin \theta}

iv. limxπsinθπx\quad \lim _{x \rightarrow \pi} \frac{\sin \theta}{\pi-x}

v. limx0sinaxsinbx\quad \lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}

vi. limx0xtanx\quad \lim _{x \rightarrow 0} \frac{x}{\tan x}

vii. limx01cos2xx2\quad \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^{2}}

viii. limx01cosxsin2x\quad \lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^{2} x}

ix. limθ0sin2θθ\quad \lim _{\theta \rightarrow 0} \frac{\sin ^{2} \theta}{\theta}

x. limx0secxcosxx\quad \lim _{x \rightarrow 0} \frac{\sec x-\cos x}{x}

xi. limθ01cospθ1cosqθ\quad \lim _{\theta \rightarrow 0} \frac{1-\cos p \theta}{1-\cos q \theta}

xii. limθ0tanθsinθsin3θ\quad \lim _{\theta \rightarrow 0} \frac{\tan \theta-\sin \theta}{\sin ^{3} \theta}

Background and Explanation

The problem involves evaluating several limits, primarily related to trigonometric functions. Some key concepts include:

  • Trigonometric Limits: Basic limits of sine and cosine functions as x0x \to 0.
  • Limit Laws: Using properties such as the limit of sine and cosine functions and using substitution methods to simplify the expressions.
  • Algebraic Manipulations: Simplifying expressions through common techniques like factoring, multiplying by conjugates, or using standard limit properties.

Solution

i. limh0sin7xx\lim _{h \rightarrow 0} \frac{\sin 7 x}{x}

We can rewrite this as:

limh0sin7x7x7=7limh0sin7x7x=7(1)=7\lim _{h \rightarrow 0} \frac{\sin 7 x}{\frac{7 x}{7}} = 7 \lim _{h \rightarrow 0} \frac{\sin 7 x}{7 x} = 7(1) = 7

ii. limx0sinx0x\lim _{x \rightarrow 0} \frac{\sin x^{0}}{x}

Since x0=1x^0 = 1 for any xx:

=limx0sinxπ180xπ180π180=limx0sinxπ180xπ180π180=1π180=π180= \lim _{x \rightarrow 0} \frac{\sin \frac{x \pi}{180}}{x} \cdot \frac{\frac{\pi}{180}}{\frac{\pi}{180}} = \lim _{x \rightarrow 0} \frac{\sin \frac{x \pi}{180}}{\frac{x \pi}{180}} \cdot \frac{\pi}{180} = 1 \cdot \frac{\pi}{180} = \frac{\pi}{180}

iii. limx01cosθsinθ\lim _{x \rightarrow 0} \frac{1-\cos \theta}{\sin \theta}

We multiply by the conjugate:

limx01cosθsinθ1+cosθ1+cosθ=limx0sin2θsinθ(1+cosθ)=limx0sinθ1+cosθ=02=0\lim _{x \rightarrow 0} \frac{1-\cos \theta}{\sin \theta} \cdot \frac{1+\cos \theta}{1+\cos \theta} = \lim _{x \rightarrow 0} \frac{\sin ^2 \theta}{\sin \theta (1+\cos \theta)} = \lim _{x \rightarrow 0} \frac{\sin \theta}{1+\cos \theta} = \frac{0}{2} = 0

iv. limxπsinθπx\lim _{x \rightarrow \pi} \frac{\sin \theta}{\pi-x}

Let x=πθx = \pi - \theta so that as xπx \to \pi, θ0\theta \to 0:

=limθ0sin(πθ)π(πθ)=limθ0sinθθ=1= \lim _{\theta \rightarrow 0} \frac{\sin (\pi - \theta)}{\pi - (\pi - \theta)} = \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1

v. limx0sinaxsinbx\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}

This is simplified as:

=limx0(ab)=ab= \lim _{x \rightarrow 0} \left( \frac{a}{b} \right) = \frac{a}{b}

vi. limx0xtanx\lim _{x \rightarrow 0} \frac{x}{\tan x}

Using the fact that limx0sinxx=1\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1:

=limx0cosxsinxx=1= \lim _{x \rightarrow 0} \frac{\cos x}{\frac{\sin x}{x}} = 1

vii. limx01cos2xx2\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^2}

Using the identity cos2x=2cos2x1\cos 2x = 2\cos^2 x - 1:

=limx02sin2xx2=2limx0sinxx=2(1)=2= \lim _{x \rightarrow 0} \frac{2 \sin^2 x}{x^2} = 2 \cdot \lim _{x \rightarrow 0} \frac{\sin x}{x} = 2(1) = 2

viii. limx01cosxsin2x\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin^2 x}

We multiply by the conjugate:

=limx0sin2xsin2x(1+cosx)=12= \lim _{x \rightarrow 0} \frac{\sin^2 x}{\sin^2 x (1+\cos x)} = \frac{1}{2}

ix. limθ0sin2θθ\lim _{\theta \rightarrow 0} \frac{\sin^2 \theta}{\theta}

Simplifying:

=limθ0sin2θθθ=0= \lim _{\theta \rightarrow 0} \frac{\sin^2 \theta}{\theta} \cdot \theta = 0

x. limx0secxcosxx\lim _{x \rightarrow 0} \frac{\sec x - \cos x}{x}

Simplifying the expression:

=limx01cos2xxcosx=0= \lim _{x \rightarrow 0} \frac{1-\cos^2 x}{x \cos x} = 0

xi. limθ01cospθ1cosqθ\lim _{\theta \rightarrow 0} \frac{1-\cos p \theta}{1-\cos q \theta}

Using the small angle approximations:

=p2q2= \frac{p^2}{q^2}

xii. limθ0tanθsinθsin3θ\lim _{\theta \rightarrow 0} \frac{\tan \theta - \sin \theta}{\sin^3 \theta}

Simplifying:

=12= \frac{1}{2}

Key Formulas or Methods Used

  • Limit of Sine and Cosine: limx0sinxx=1\lim_{x \to 0} \frac{\sin x}{x} = 1
  • Conjugate Multiplication: Used to simplify trigonometric expressions.
  • Small Angle Approximation: Simplifications for sinxx\sin x \approx x and cosx1\cos x \approx 1 as x0x \to 0.

Summary of Steps

  1. For each limit, identify trigonometric identities or simplifications (like sinxx\sin x \approx x as x0x \to 0).
  2. Simplify the expression using known limits or algebraic manipulation (e.g., conjugate multiplication, factoring).
  3. Apply the standard limits of trigonometric functions to solve for the final result.
  4. Ensure all intermediate steps are logically connected to avoid skipping key steps.

For better understanding refer to video provided on 03_Ex 1.3