Question

find the value of limit of X tends to infinity of x square Sin of 1 by x minus x upon 1 minus Mod of x​

Answer 1

Answer:

0

Step-by-step explanation:

Answer 2

Concept Introduction:-

Mathematics defines a limit as the value that have a functional approaches when the input approaches to a certain value.

Given Information:-

We have been given that limit of $X$ tends to infinity of $x$ square Sin of $1$ by $x$ minus $x$ upon $1$ minus Mod of $x$

To Find:-

We have to find that the value of limit of $X$ tends to infinity of $x$ square Sin of $1$ by $x$ minus $x$ upon $1$ minus Mod of $x$

Solution:-

According to the problem

$\lim _{x \rightarrow \infty}\left[\frac{x^{2} \sin \left(\frac{1}{x}\right)-x}{1-x \mid}\right] =\lim _{x \rightarrow \infty}\left[\frac{x^{2} \sin \left(\frac{1}{x}\right)-x}{1-x}\right]\\=\lim _{x \rightarrow \infty} \frac{\frac{\sin \left(x^{-1}\right)-1}{x^{-1}}}{x^{-1}-1} \\ =\lim _{y \rightarrow 0} \frac{\frac{\sin y}{y}-1}{y-1}=\frac{1-1}{0-1}=0$

Final Answer:-

The value of limit of $X$ tends to infinity of $x$ square Sin of $1$ by $x$ minus $x$ upon $1$ minus Mod of $x$​ is $0$.