mathop {lim }limits_{x to 0} left( frac{x}{ta | vedclass

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

Question

(B) Let $	an^{-1} 2x = 	heta$. Then $2x = 	an 	heta$,which implies $x = \frac{1}{2} 	an 	heta$. As $x 	o 0$,$	heta 	o 0$. Substituting these

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Let $	an^{-1} 2x = 	heta$. Then $2x = 	an 	heta$,which implies $x = \frac{1}{2} 	an 	heta$. As $x 	o 0$,$	heta 	o 0$. Substituting these into the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{x}{	an^{-1} 2x} = \mathop {\lim }\limits_{	heta 	o 0} \frac{\frac{1}{2} 	an 	heta}{	heta} = \frac{1}{2} \mathop {\lim }\limits_{	heta 	o 0} \frac{	an 	heta}{	heta}$. Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{	an 	heta}{	heta} = 1$,the result is $\frac{1}{2} 	imes 1 = \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\tan^{-1} 2x} \right) = $

Similar Questions

If $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$,the equation whose roots are $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ is

If $0 < x < \frac{\pi}{2}$,then

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {\frac{1}{2}(1 - \cos 2x)} }}{x} = $

If $\Delta(x) = \begin{vmatrix} e^x & -1 \\ \sin x - 1 & 1 \end{vmatrix}$,then $\lim_{x \rightarrow 0} \frac{\Delta(x)}{x} = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module