lim _{x rightarrow 0}left[tan left(x+frac{pi} | vedclass

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

Question

(A) Let $L = \lim _{x ightarrow 0}\left[	an \left(x+\frac{\pi}{4}ight)ight]^{1 / x}$.Since this is of the form $1^\infty$,we use the formula $\

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(A) Let $L = \lim _{x ightarrow 0}\left[	an \left(x+\frac{\pi}{4}ight)ight]^{1 / x}$.Since this is of the form $1^\infty$,we use the formula $\lim_{x 	o a} [f(x)]^{g(x)} = e^{\lim_{x 	o a} [f(x)-1]g(x)}$.$L = e^{\lim _{x ightarrow 0}\left[	an \left(x+\frac{\pi}{4}ight)-1ight] \frac{1}{x}}$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have $	an(x+\frac{\pi}{4}) = \frac{	an x + 1}{1 - 	an x}$.$L = e^{\lim _{x ightarrow 0}\left[\frac{	an x+1}{1-	an x}-1ight] \frac{1}{x}} = e^{\lim _{x ightarrow 0} \left[\frac{	an x+1-1+	an x}{1-	an x}ight] \frac{1}{x}}$.$L = e^{\lim _{x ightarrow 0} \frac{2 	an x}{x(1-	an x)}}$.Since $\lim_{x 	o 0} \frac{	an x}{x} = 1$ and $\lim_{x 	o 0} (1-	an x) = 1$,we get $L = e^{2 	imes 1 / 1} = e^2$.

$\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$ is equal to

Similar Questions

Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$,$l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$ and $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$,then:

$\lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right) = $

Let $f(x) = \lim_{y \to 0} \frac{(1 - \cos(xy))\tan(xy)}{y^3}$. Then the number of solutions of the equation $f(x) = \sin x, x \in R$ is:

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

$\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module