If f(x) = left(frac{1+tan x}{1+sin x}right)^{ | vedclass

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

Question

(B) For $f(x)$ to be continuous at $x=0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \left(\frac{1+	an x}{1+\sin x}ight)^{\operatorna

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) For $f(x)$ to be continuous at $x=0$,we must have $f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \left(\frac{1+	an x}{1+\sin x}ight)^{\operatorname{cosec} x} = \lim_{x 	o 0} \exp\left(\operatorname{cosec} x \cdot \ln\left(\frac{1+	an x}{1+\sin x}ight)ight)$.Using the expansion $	an x \approx x + \frac{x^3}{3}$ and $\sin x \approx x - \frac{x^3}{6}$,we have:$\frac{1+	an x}{1+\sin x} \approx \frac{1+x}{1+x} = 1$.Since the limit is of the form $1^\infty$,we use $\lim_{x 	o 0} (1+u)^v = e^{\lim_{x 	o 0} uv}$.$\lim_{x 	o 0} \frac{1}{\sin x} \left(\frac{1+	an x}{1+\sin x} - 1ight) = \lim_{x 	o 0} \frac{1}{\sin x} \left(\frac{	an x - \sin x}{1+\sin x}ight) = \lim_{x 	o 0} \frac{	an x - \sin x}{\sin x(1+\sin x)}$.Using $	an x - \sin x = \sin x(\sec x - 1) \approx x \cdot \frac{x^2}{2} = \frac{x^3}{2}$ and $\sin x(1+\sin x) \approx x(1+x) \approx x$,the limit is $\lim_{x 	o 0} \frac{x^3/2}{x} = 0$.Thus,$f(0) = e^0 = 1$.

If $f(x) = \left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$ is continuous at $x=0$,then $f(0)$ is equal to

Similar Questions

If the function $f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$,$x \in R$,is continuous at $x = 0$,then $f(0)$ is equal to:

If $f: R \rightarrow R$ is defined by $f(x) = x - [x]$,where $[x]$ is the greatest integer not exceeding $x$,then the set of points of discontinuity of $f$ is

Which one of the following functions is not continuous on $(0, \pi )$?

If $f(x) = \begin{cases} \sin^{-1}|x|, & x \ne 0 \\ 0, & x = 0 \end{cases}$ then

Let $a, b, c$ be three real numbers. If the function $f(x) = \begin{cases} \cos(2x + \pi) & \text{if } x \leq 0 \\ ax^2 + b & \text{if } 0 < x < 1 \\ cx + 4 & \text{if } 1 \leq x \leq 2 \\ 3a + 1 & \text{if } x \geq 2 \end{cases}$ is continuous everywhere,then $b^2 - bc + c^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module