lim _{x rightarrow 2}(x-1)^{frac{1}{3x-6}} = | vedclass

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

Question

(C) माना $L = \lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}}$.चूंकि यह $1^{\infty}$ का रूप है,हम सूत्र $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _

Vedclass · Accepted Answer

$e^{\frac{1}{3}}$

Vedclass · Answer

(C) माना $L = \lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}}$.चूंकि यह $1^{\infty}$ का रूप है,हम सूत्र $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} g(x)(f(x)-1)}$ का उपयोग करते हैं।यहाँ,$f(x) = x-1$ और $g(x) = \frac{1}{3x-6}$ है।$L = e^{\lim _{x \rightarrow 2} \frac{1}{3(x-2)} (x-1-1)}$.$L = e^{\lim _{x \rightarrow 2} \frac{x-2}{3(x-2)}}$.$L = e^{\lim _{x \rightarrow 2} \frac{1}{3}} = e^{\frac{1}{3}}$.

$\lim _{x \rightarrow 2}(x-1)^{\frac{1}{3x-6}} = $

Similar Questions

$\lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3} = $

$\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}}$ का मान क्या है?

$\lim _{x \rightarrow 0} \left[ \frac{1}{x} - \frac{1}{e^x - 1} \right] = $

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ का मान ज्ञात कीजिए।

$\mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module