lim _{x rightarrow infty}left(frac{x^{2}-2 x+ | vedclass

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

Question

(A) આપેલ લક્ષ $1^{\infty}$ સ્વરૂપમાં છે.આપણે સૂત્ર $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$ નો ઉપયોગ કરીએ છી

Vedclass · Accepted Answer

$e^{2}$

Vedclass · Answer

(A) આપેલ લક્ષ $1^{\infty}$ સ્વરૂપમાં છે.આપણે સૂત્ર $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$ નો ઉપયોગ કરીએ છીએ.$\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x} = e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2} - 1\right)}$$= e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1 - (x^{2}-4 x+2)}{x^{2}-4 x+2}\right)}$$= e^{\lim _{x \rightarrow \infty} x \left(\frac{2 x-1}{x^{2}-4 x+2}\right)}$$= e^{\lim _{x \rightarrow \infty} \frac{2 x^{2}-x}{x^{2}-4 x+2}}$અંશ અને છેદને $x^{2}$ વડે ભાગતા,આપણને મળે છે:$= e^{\lim _{x \rightarrow \infty} \frac{2 - 1/x}{1 - 4/x + 2/x^{2}}} = e^{2/1} = e^{2}$.

$\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x}$ ની કિંમત શોધો.

Similar Questions

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$ ની કિંમત શોધો.

$\lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} = $

જો $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$ હોય,તો $\frac{\log _e \alpha}{1+\log _e \alpha}$ ની કિંમત શોધો:

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module