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 $\mathop {\lim }\limits_{x \to 0} \frac{(1 - \cos 2x)\sin 5x}{x^2 \sin 3x}$ is

The value of the limit $\lim _{\theta \rightarrow 0} \frac{\tan (\pi \cos ^{2} \theta)}{\sin (2 \pi \sin ^{2} \theta)}$ is equal to :

If $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$,then $\lim_{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)} = $

$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module