mathop {lim }limits_{x to 0} left( {frac{{tan | vedclass

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

Question

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left( {\frac{{	an 3x}}{x} + \cos x} ight)$.Using the property of limits,$\mat

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \left( {\frac{{	an 3x}}{x} + \cos x} ight)$.Using the property of limits,$\mathop {\lim }\limits_{x 	o a} [f(x) + g(x)] = \mathop {\lim }\limits_{x 	o a} f(x) + \mathop {\lim }\limits_{x 	o a} g(x)$,we get:$\mathop {\lim }\limits_{x 	o 0} \frac{{	an 3x}}{x} + \mathop {\lim }\limits_{x 	o 0} \cos x$.For the first term,multiply and divide by $3$:$\mathop {\lim }\limits_{x 	o 0} 3 \cdot \frac{{	an 3x}}{{3x}} + \mathop {\lim }\limits_{x 	o 0} \cos x$.Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{{	an 	heta}}{	heta} = 1$,the first term becomes $3 \cdot 1 = 3$.For the second term,$\mathop {\lim }\limits_{x 	o 0} \cos x = \cos(0) = 1$.Therefore,the total limit is $3 + 1 = 4$.

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan 3x}}{x} + \cos x} \right) = $

Similar Questions

Find the value of $\lim _{x \rightarrow 0} \frac{\sin (x^m)}{(\sin x)^n}$,given that $n < m$.

The value of $\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ is:

$\lim _{x \rightarrow 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{\left( {1 - \cos 2x} \right)}^2}}}$ equals

$\mathop {\lim }\limits_{x \to 0} \frac{{3\sin x - \sin 3x}}{{{x^3}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module