The value of lim_{x to 3} frac{[x] - 3}{x - 3 | vedclass

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

Question

(D) To find the limit $\lim_{x 	o 3} \frac{[x] - 3}{x - 3}$, we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.Step $1$: Calcu

Vedclass · Accepted Answer

does not exist

Vedclass · Answer

(D) To find the limit $\lim_{x 	o 3} \frac{[x] - 3}{x - 3}$, we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.Step $1$: Calculate $LHL$ as $x 	o 3^-$. For $x$ slightly less than $3$, $[x] = 2$. Thus, $\lim_{x 	o 3^-} \frac{2 - 3}{x - 3} = \lim_{x 	o 3^-} \frac{-1}{x - 3}$. As $x 	o 3^-$, $(x - 3)$ is a very small negative number, so $\frac{-1}{x - 3} 	o \infty$.Step $2$: Calculate $RHL$ as $x 	o 3^+$. For $x$ slightly greater than or equal to $3$, $[x] = 3$. Thus, $\lim_{x 	o 3^+} \frac{3 - 3}{x - 3} = \lim_{x 	o 3^+} \frac{0}{x - 3} = 0$.Step $3$: Since the $LHL$ $(\infty)$ is not equal to the $RHL$ $(0)$, the limit does not exist.

The value of $\lim_{x \to 3} \frac{[x] - 3}{x - 3}$, where $[\cdot]$ denotes the greatest integer function, is...

Similar Questions

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

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

The limit of the perimeter of the regular $n$-gons inscribed in a circle of radius $R$ as $n \to \infty$ is

$\lim _{n \rightarrow \infty} \frac{\left[6^2+12^2+18^2+\ldots+(6 n)^2\right]^2}{[5+10+15+\ldots+5 n]\left[2^3+4^3+6^3+\ldots+(2 n)^3\right]} =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module