Evaluation of Trigonometric Limits

Evaluation of Trigonometric Limits

 \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1\) (where θ is in radians).

Proof:

1. Consider a circle of radius such that ∠AOB = θ, where θ is measured in radius and its value is very small. Suppose the tangent at A meets. OB produced at P. from Fig. we have

Area of ΔABC < area of sector OAB < Area of ΔOAP

½ x OA x OB sinθ < ½ (OA)² θ < ½ OA x AP.

½ x r² sinθ < ½ r²θ < ½ r² tanθ

Sinθ < θ < tanθ

1 < θ/sinθ < 1/cosθ

1 > sinθ/θ > cosθ

i.e., \(1>\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }>\underset{x\to 0}{\mathop{\lim }}\,\cos \theta \).

(or)

\(\underset{x\to 0}{\mathop{\lim }}\,\cos \theta <\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }<1\).

\(1<\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }<1\).

By sandwich theorem

\(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1\).

2. \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan \theta }{\theta }=1\).

 We have \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\tan \theta }{\theta }\).

\(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta /\cos \theta }{\theta }\).

\(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }.\underset{x\to 0}{\mathop{\lim }}\,\cos \theta \).

Since \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1\).

= (1) (1) = 1

3. \(\underset{x\to a}{\mathop{\lim }}\,\frac{\sin (\theta -a)}{(\theta -a)}=1\).

We have

\(\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin (a+h-a)}{(a+h-a)}\).

 \(\underset{h\to 0}{\mathop{\lim }}\,\frac{\sin (h)}{h}\).

Since \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1\).

= 1

4. \(\underset{x\to a}{\mathop{\lim }}\,\frac{\tan (\theta -a)}{(\theta -a)}=1\).

5. \(\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\sin }^{-1}}x}{x}=1\).

6. \(\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\operatorname{Tan}}^{-1}}x}{x}=1\).

Example: Evaluate the following \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{Sin}3x}{x}\).

Solution: Given that \(\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{Sin}3x}{x}\).

\(\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{Sin}3x}{x}\times \frac{3}{3}\).

\(\underset{x\to 0}{\mathop{\lim }}\,3.\frac{\operatorname{Sin}3x}{3x}\).

\(3.\underset{x\to 0}{\mathop{\lim }}\,\frac{\operatorname{Sin}3x}{3x}\).

= 3 (1) = 3