lim _{x rightarrow 0} frac{2^{2 x}-2^{x+1}+2- | vedclass

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

Question

(B) આપેલ છે,$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}$ $= \lim _{x \rightarrow 0} \frac{(2^x)^2 - 2 \cdot 2^x + 1 + 1 - \cos 2x}

Vedclass · Accepted Answer

$2+(\log _e 2)^2$

Vedclass · Answer

(B) આપેલ છે,$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}$ $= \lim _{x \rightarrow 0} \frac{(2^x)^2 - 2 \cdot 2^x + 1 + 1 - \cos 2x}{x^2}$ $= \lim _{x \rightarrow 0} \frac{(2^x - 1)^2 + (1 - \cos 2x)}{x^2}$ $= \lim _{x \rightarrow 0} \left( \frac{2^x - 1}{x} \right)^2 + \lim _{x \rightarrow 0} \frac{1 - \cos 2x}{x^2}$ પ્રમાણિત લક્ષ $\lim _{x \rightarrow 0} \frac{a^x - 1}{x} = \log _e a$ અને $\lim _{x \rightarrow 0} \frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$ નો ઉપયોગ કરતા: $= (\log _e 2)^2 + \frac{2^2}{2}$ $= (\log _e 2)^2 + 2$

$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2} = $

Similar Questions

ધારો કે $f(x) = \frac{x - 1}{2x^2 - 7x + 5}$. તો:

લક્ષ $\lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$

$\mathop {\lim }\limits_{x \to 1} \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3} = $

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module