Evaluate the given limit: mathop {lim }limits | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{\sin ax}{\sin bx}$.At $x=0$,the expression takes the indeterminate form $\frac{0}{

Vedclass · Accepted Answer

$\frac{a}{b}$

Vedclass · Answer

(A) We are given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{\sin ax}{\sin bx}$.At $x=0$,the expression takes the indeterminate form $\frac{0}{0}$.We use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$.$\mathop {\lim }\limits_{x 	o 0} \frac{\sin ax}{\sin bx} = \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin ax}{ax} \cdot ax \cdot \frac{1}{\frac{\sin bx}{bx} \cdot bx} ight)$$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin ax}{ax} ight) \cdot \frac{1}{\mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin bx}{bx} ight)} \cdot \frac{ax}{bx}$$= 1 \cdot \frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}$.

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx}$,where $a, b \neq 0$.

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin (x + a) + \sin (a - x) - 2\sin a}}{{x\sin x}}} \right] = $

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos mx}}{{1 - \cos nx}} = $

$\lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}}$ is equal to :

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x + \sin x}}{x} = $

$\lim _{x \rightarrow 0} \left( \frac{\sin (\pi \cos ^2 x)}{x^2} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module