mathop {lim }limits_{x to 0} frac{{log cos x} | vedclass

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

Question

(A) $\frac{0}{0}$ સ્વરૂપ હોવાથી $L'	ext{Hospital's rule}$ નો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{{\log \cos x}}{x} = \mathop {\lim }\l

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) $\frac{0}{0}$ સ્વરૂપ હોવાથી $L'	ext{Hospital's rule}$ નો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{{\log \cos x}}{x} = \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{d}{{dx}}(\log \cos x)}}{{\frac{d}{{dx}}(x)}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{\frac{1}{{\cos x}} \cdot (-\sin x)}}{1}$$= \mathop {\lim }\limits_{x 	o 0} (-	an x)$$= -	an(0) = 0$

$\mathop {\lim }\limits_{x \to 0} \frac{{\log \cos x}}{x} = $

Similar Questions

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ ની કિંમત શોધો.

લક્ષની કિંમત શોધો: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$

$\mathop {\lim }\limits_{x \to 0} \frac{{{{27}^x} - {9^x} - {3^x} + 1}}{{\sqrt 5 - \sqrt {4 + \cos x} }}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to \alpha } \frac{{\sin x - \sin \alpha }}{{x - \alpha }} = $

જો $f(5)=7$ અને $f'(5)=7$ હોય, તો $\lim_{x \rightarrow 5} \frac{x f(5)-5 f(x)}{x-5}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module