lim _{x rightarrow 1}(1-x) tan left(frac{pi}{ | vedclass

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

Question

(B) Let $h = 1-x$. As $x ightarrow 1$,$h ightarrow 0$,so $x = 1-h$. Substituting this into the limit: $\lim _{h ightarrow 0} h 	an \left(\frac{

Vedclass · Accepted Answer

$\frac{2}{\pi}$

Vedclass · Answer

(B) Let $h = 1-x$. As $x ightarrow 1$,$h ightarrow 0$,so $x = 1-h$. Substituting this into the limit: $\lim _{h ightarrow 0} h 	an \left(\frac{\pi}{2}(1-h)ight) = \lim _{h ightarrow 0} h 	an \left(\frac{\pi}{2} - \frac{\pi h}{2}ight)$ Since $	an(\frac{\pi}{2} - 	heta) = \cot(	heta)$,we have: $\lim _{h ightarrow 0} h \cot \left(\frac{\pi h}{2}ight) = \lim _{h ightarrow 0} h \frac{1}{	an \left(\frac{\pi h}{2}ight)}$ Multiply and divide by $\frac{\pi}{2}$: $\lim _{h ightarrow 0} \frac{h \cdot \frac{\pi}{2}}{	an \left(\frac{\pi h}{2}ight)} \cdot \frac{2}{\pi} = 1 \cdot \frac{2}{\pi} = \frac{2}{\pi}$.

$\lim _{x \rightarrow 1}(1-x) \tan \left(\frac{\pi}{2} x\right) = $

Similar Questions

$\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$ is equal to

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

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = $

$\mathop {\lim }\limits_{x \to 0} \,\frac{{x\,\cot \,\left( {4x} \right)}}{{{{\sin }^2}\,x\,{{\cot }^2}\,\left( {2x} \right)}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module