lim _{x rightarrow 0} frac{27^x-9^x-3^x+1}{sq | vedclass

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

Question

(C) ધારો કે $L = \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$.અંશના અવયવ પાડતા: $27^x-9^x-3^x+1 = (9^x-1)(3^x-1)$.છેદનું સ

Vedclass · Accepted Answer

$8 \sqrt{5}(\log 3)^2$

Vedclass · Answer

(C) ધારો કે $L = \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$.અંશના અવયવ પાડતા: $27^x-9^x-3^x+1 = (9^x-1)(3^x-1)$.છેદનું સંમેયીકરણ કરતા: $\frac{1}{\sqrt{5}-\sqrt{4+\cos x}} = \frac{\sqrt{5}+\sqrt{4+\cos x}}{1-\cos x}$.તેથી,$L = \lim _{x \rightarrow 0} \frac{(9^x-1)(3^x-1)}{1-\cos x} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.$1-\cos x = 2 \sin^2(x/2)$ નો ઉપયોગ કરતા,$L = \lim _{x \rightarrow 0} \frac{(9^x-1)(3^x-1)}{2 \sin^2(x/2)} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.$x^2$ વડે ગુણતા અને ભાગતા: $L = \lim _{x}$ ${\rightarrow 0} \frac{(\frac{9^x-1}{x})(\frac{3^x-1}{x})}{2 \cdot (\frac{\sin(x/2)}{x/2})^2 \cdot (1/4)} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.$L = \frac{\ln 9 \cdot \ln 3}{1/2} \cdot 2 \sqrt{5} = 8 \sqrt{5} (\ln 3)^2$.

$\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=$

Similar Questions

લક્ષની કિંમત શોધો: $\mathop {\lim }\limits_{x \to 3} (x+3)$

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

જો $0 \leq x \leq \pi / 2$ હોય,તો $\lim _{x \rightarrow a} \frac{|2 \cos x-1|}{2 \cos x-1}$

$\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right) = $

જો $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,અને $S_3 = \sum_{r=1}^{n} r^3$ હોય,તો $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module