If l = lim_{x rightarrow 0} frac{x}{|x| + x^2 | vedclass

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

Question

(D) Let $f(x) = \frac{x}{|x| + x^2}$.For the left-hand limit as $x \rightarrow 0^-$,we have $|x| = -x$:$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \right

Vedclass · Accepted Answer

non-existent

Vedclass · Answer

(D) Let $f(x) = \frac{x}{|x| + x^2}$.For the left-hand limit as $x ightarrow 0^-$,we have $|x| = -x$:$\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0} \frac{x}{-x + x^2} = \lim_{x ightarrow 0} \frac{1}{-1 + x} = -1$.For the right-hand limit as $x ightarrow 0^+$,we have $|x| = x$:$\lim_{x ightarrow 0^+} f(x) = \lim_{x ightarrow 0} \frac{x}{x + x^2} = \lim_{x ightarrow 0} \frac{1}{1 + x} = 1$.Since $\lim_{x ightarrow 0^-} f(x) 
eq \lim_{x ightarrow 0^+} f(x)$,the limit does not exist.

If $l = \lim_{x \rightarrow 0} \frac{x}{|x| + x^2}$,then the value of $l$ is

Similar Questions

$\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=$

$\lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right]$ is equal to

The value of $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$ is

$\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2+x^3}$ is equal to

The value of $\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{x^2} + bx + 4}}{{{x^2} + ax + 5}}} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module