The limit of sine of angle $x$ minus $1$ by $x$ square minus $1$ as the value of $x$ approaches to $1$ is indeterminate as per the direct substitution method.

$\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\sin{(x-1)}}{x^2-1}}$ $\,=\,$ $\dfrac{0}{0}$

It clears that it is not possible to find the limit of sin of $x$ minus $1$ by $x$ squared minus $1$ as the value $x$ is closer to $1$ as per the direct substitution. Due to the indeterminate form, the limit of the sin of angle $x$ minus $1$ by $x$ square minus $1$ as $x$ tends to $1$ can be calculated by the L’Hospital’s rule.

In the rational function, $\sin{(x-1)}$ and $x^2-1$ are the trigonometric and algebraic functions respectively and they both are defined in $x$. So, the functions $\sin{(x-1)}$ and $x^2-1$ should be differentiated with respect to $x$ for applying the L’Hospital’s Rule.

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

It is time to find the derivative of a trigonometric function $\sin{(x-1)}$ and an algebraic function $x^2-1$ with respect to $x$.

Firstly, look at the numerator. $\sin{(x-1)}$ is a trigonometric function but its angle is in the form of an algebraic function and it clears that $\sin{(x-1)}$ is a composition of both trigonometric and algebraic functions. The derivative of composite function $\sin{(x-1)}$ can be calculated as per the chain rule.

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

Look at the second factor in the numerator, the algebraic function $x-1$ represents the subtraction of two functions $x$ and $1$. Hence, its derivative can be calculated by the subtraction rule of differentiation.

In the denominator, the algebraic function $x^2-1$ represents the difference of two functions $x^2$ and $1$. So, the derivative of $x^2-1$ can be calculated as per the difference rule of the derivatives.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\dfrac{d}{d(x-1)}\,\sin{(x-1)} \times \Big(\dfrac{d}{dx}\,(x)-\dfrac{d}{dx}\,(1)\Big)}{\dfrac{d}{dx}\,(x^2)-\dfrac{d}{dx}\,(1)}}$

Now, let’s find the derivative of every function. According to the derivative rule of sine function, the derivative of $\sin{(x-1)}$ with respect to $x-1$ is $\cos{(x-1)}$. According to the derivative rule of a variable, the derivative of $x$ with respect to same variable is one. According to the derivative rule of a constant, the derivative of one with respect to $x$ is zero. As per the power rule of derivatives, the derivative of $x$ square with respect to $x$ is $2$ times $x$.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\cos{(x-1)} \times (1-0)}{2x^{2-1}-0}}$

It is time to simplify the whole function for preparing the function to calculate its limit.

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

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\cos{(x-1)} \times 1}{2x}}$

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

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \dfrac{\cos{(x-1)}}{2 \times x}}$

After applying the L’Hopital’s Rule, the limit of sine of angle $x$ minus $1$ by $x$ square minus $1$ as $x$ approaches $1$ is transformed as the limit of cosine of angle $x$ minus $1$ by $2$ times $x$ as the value of $x$ is closer to $1$. Now, let’s try to find the limit of the function by the direct substitution.

$=\,\,\,$ $\dfrac{\cos{(1-1)}}{2 \times 1}$

$=\,\,\,$ $\dfrac{\cos{(0)}}{2}$

$=\,\,\,$ $\dfrac{\cos{0}}{2}$

According to the trigonometry, the cosine of angle zero radian is one. So, substitute its value in the numerator to finish the process of finding the limit of the given function.

$=\,\,\,$ $\dfrac{1}{2}$

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