Let [x] denote the greatest integer less than | vedclass

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

Question

(B) Given the limit $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.As $x \rightarrow 2^{+}$,$x$ is slightly grea

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(B) Given the limit $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.As $x \rightarrow 2^{+}$,$x$ is slightly greater than $2$,so $[x] = 2$.Also,as $x \rightarrow 2^{+}$,$\frac{x}{3}$ is slightly greater than $\frac{2}{3}$,so $\left[\frac{x}{3}\right] = 0$.Substituting these values into the expression:$\lim _{x}$ ${\rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right) = \frac{2^3}{3} - 0^3 = \frac{8}{3} - 0 = \frac{8}{3}$.

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)=$

Similar Questions

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

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

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$

$\lim _{x \rightarrow 0} \left( \frac{\tan x}{\sqrt{2x+4}-2} \right)$ is equal to

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module