If lim_{x to 0} frac{(4^x - 1)^3}{tan(frac{x} | vedclass

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

Question

(QUE_A) We use the standard limits: $\lim_{x 	o 0} \frac{a^x - 1}{x} = \log a$, $\lim_{x 	o 0} \frac{	an x}{x} = 1$, and $\lim_{x 	o 0} \frac{\log

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(QUE_A) We use the standard limits: $\lim_{x 	o 0} \frac{a^x - 1}{x} = \log a$, $\lim_{x 	o 0} \frac{	an x}{x} = 1$, and $\lim_{x 	o 0} \frac{\log(1+x)}{x} = 1$.Dividing numerator and denominator by $x^3$, we get:$\lim_{x 	o 0} \frac{\left(\frac{4^x - 1}{x}ight)^3}{\frac{	an(x/4)}{x} \cdot \frac{\log(1 + x^2/3)}{x^2}} = \frac{(\log 4)^3}{\frac{1}{4} \cdot \frac{1}{3}} = 12 (\log 4)^3$.Given $12 (\log 4)^3 = 96 (\log a)^b$, we can simplify this as:$12 (2 \log 2)^3 = 96 (\log a)^b$$12 \cdot 8 (\log 2)^3 = 96 (\log a)^b$$96 (\log 2)^3 = 96 (\log a)^b$Comparing both sides, we get $a = 2$ and $b = 3$.Therefore, $a + b = 2 + 3 = 5$.

If $\lim_{x \to 0} \frac{(4^x - 1)^3}{\tan(\frac{x}{4}) \log(1 + \frac{x^2}{3})} = 96(\log a)^b$, then $(a + b) = $

Similar Questions

Evaluate the limit: $\mathop {\lim }\limits_{x \to \infty } [x({a^{1/x}} - 1)]$,where $a > 1$.

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

$\lim _{x \rightarrow 2}\left(\sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}\right)$ is equal to :

$\lim _{x \rightarrow 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2} = $

If $\mathop {\lim }\limits_{n \to \infty } n \cos \left( \frac{\pi }{4n} \right) \sin \left( \frac{\pi }{4n} \right) = k$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module