Evaluate the given limit: mathop {lim }limits | vedclass

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

Question

(B) Given limit: $\mathop {\lim }\limits_{x 	o -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$At $x = -2$,the expression takes the indeterminate form $\

Vedclass · Accepted Answer

$-\frac{1}{4}$

Vedclass · Answer

(B) Given limit: $\mathop {\lim }\limits_{x 	o -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$At $x = -2$,the expression takes the indeterminate form $\frac{0}{0}$.Simplify the numerator:$\frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x}$Now substitute this back into the limit:$\mathop {\lim }\limits_{x 	o -2} \frac{\frac{2 + x}{2x}}{x + 2} = \mathop {\lim }\limits_{x 	o -2} \frac{2 + x}{2x(x + 2)}$Cancel the common factor $(x + 2)$:$= \mathop {\lim }\limits_{x 	o -2} \frac{1}{2x}$Evaluate the limit by substituting $x = -2$:$= \frac{1}{2(-2)} = -\frac{1}{4}$

Evaluate the given limit: $\mathop {\lim }\limits_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

Similar Questions

The limit of the interior angle of a regular polygon of $n$ sides as $n \rightarrow \infty$ is

If $\lim _{x \rightarrow 0}\left(\frac{1+cx}{1-cx}\right)^{1/x}=4$, then $\lim _{x \rightarrow 0}\left(\frac{1+2cx}{1-2cx}\right)^{1/x}$ is

Let $f(x) = \frac{x \cdot 2^x - x}{1 - \cos x}$ and $g(x) = 2^x \sin \left( \frac{\ln 2}{2^x} \right)$,then:

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

Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{{y^2}}}{x}$,where ${y^2} = ax + b{x^2} + c{x^3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module