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.

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).

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.

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