# 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.

Latest Math Topics
Jun 26, 2023
Jun 23, 2023

Latest Math Problems
Jul 01, 2023
Jun 25, 2023
###### Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Practice now

###### Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

###### Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.