Math Doubts

Evaluate $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\sin{(x-1)}}{x^2-1}}$

The limit of sine of angle $x$ minus $1$ by $x$ square minus $1$ as $x$ approaches to $1$ is written in mathematical form as follows.

$\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\sin{(x-1)}}{x^2-1}}$

Firstly, use the direct substitution to find the limit of sine of angle $x$ minus $1$ by $x$ square minus $1$ as the value of $x$ is closer to $1$.

$=\,\,\,$ $\dfrac{\sin{(1-1)}}{1^2-1}$

$=\,\,\,$ $\dfrac{\sin{(0)}}{1-1}$

$=\,\,\,$ $\dfrac{0}{0}$

It is evaluated that the limit of sine of $x$ minus $1$ by square of $x$ minus $1$ is indeterminate as the value of $x$ approaches to $1$ as per the direct substitution. So, the direct substitution method is not useful to find the limit. However, the limit of sine of angle $x$ minus one by $x$ squared minus $1$ can be calculated by the following two methods.

Factoring

Learn how to evaluate the limit of sine of $x$ minus $1$ by square of $x$ minus $1$ as the value of $x$ tends to $1$ by factorization (or factorisation).

L’Hopital’s Rule

Learn how to find the limit of sine of angle $x$ minus $1$ by $x$ square minus $1$ as the value of $x$ is closer to $1$ by using the L’Hospital’s rule.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved