lim _{x rightarrow 0} left( frac{x cdot 10^x | vedclass

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

Question

(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:$L = \li

Vedclass · Accepted Answer

$2 \log 10$

Vedclass · Answer

(B) Given limit: $L = \lim _{x \rightarrow 0} \frac{x(10^x - 1)}{1 - \cos x}$.Since this is a $\frac{0}{0}$ form,we apply $L$-Hospital's rule:$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x \ln 10 + (10^x - 1)}{\sin x}$.Again,this is a $\frac{0}{0}$ form,so we apply $L$-Hospital's rule again:$L = \lim _{x \rightarrow 0} \frac{x \cdot 10^x (\ln 10)^2 + 10^x \ln 10 + 10^x \ln 10}{\cos x}$.Substituting $x = 0$:$L = \frac{0 \cdot 10^0 (\ln 10)^2 + 10^0 \ln 10 + 10^0 \ln 10}{\cos 0} = \frac{0 + \ln 10 + \ln 10}{1} = 2 \ln 10$.Since $\log 10$ usually denotes $\log_{10} 10 = 1$ or $\ln 10$ depending on context,and given the options,the result is $2 \log 10$.

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(27 + x)}^{\frac{1}{3}}}} - 3}{{9 - {{(27 + x)}^{\frac{2}{3}}}}}$ equals.

If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ and $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$,then $12 k=$

$\lim _{x \rightarrow 2} \frac{x+3 x^2+5 x^3+7 x^4-166}{x-2} = $

The value of $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x)$ is equal to

$\mathop {\lim }\limits_{x \to 1} {\left( {\frac{4}{\pi }{{\tan }^{ - 1}}x} \right)^{\frac{1}{{({x^2} - 1)}}}}$ is equal to -

Vedclass Test Series

Exam Paper Generator

Online Exam Module