mathop {lim }limits_{x to 0} frac{{ln (cos x) | vedclass

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

Question

(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 0$.Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with res

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) The given limit is of the form $\frac{0}{0}$ as $x 	o 0$.Applying $L'Hospital's$ rule,we differentiate the numerator and the denominator with respect to $x$:$\mathop {\lim }\limits_{x 	o 0} \frac{{\ln (\cos x)}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{\cos x}} \cdot (-\sin x)}}{{2x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{-	an x}}{{2x}}$Applying $L'Hospital's$ rule again:$= \mathop {\lim }\limits_{x 	o 0} \frac{{-\sec^2 x}}{2}$$= \frac{{-1^2}}{2} = -\frac{1}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$ is equal to

Similar Questions

Evaluate the limit: $\lim _{x \rightarrow -9} \frac{(2.5)^{81-x^2}-(0.4)^{x+9}}{x+9}$

The value of $\mathop {\lim }\limits_{x \to a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$ is

$\lim _{x \rightarrow 1}\left((1-x) \tan \left(\frac{\pi x}{2}\right)\right)=$

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module