It is evaluated that the limit of sine of $x$ minus $1$ by $x$ squared minus $1$ as the value of $x$ tends to $1$ is indeterminate as per the direct substitution.

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

In fact, the limit of the quotient of $\sin{(x-1)}$ by $x^2-1$ becomes indeterminate as the value of $x$ is closer to $1$ is mainly due to the quadratic form in the algebraic function $x^2-1$.

In order to remove the indeterminate form in the given rational function, it is better to factorize (or factorise) the quadratic expression in the denominator.

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

The difference of squares of $x$ and $1$ can be factored as a product of two binomials as per the difference of squares formula.

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

Now, let’s try to factorize this rational function as a product of two fractions by the multiplication rule of fractions.

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

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

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

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \Bigg(\dfrac{\sin{(x-1)}}{(x-1)} \times \dfrac{1}{(x+1)}\Bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,1}{\normalsize \Bigg(\dfrac{\sin{(x-1)}}{x-1} \times \dfrac{1}{x+1}\Bigg)}$

According to the product rule of limits, the limit of a product of $\sin{(x-1)}$ by $x-1$ and $1$ by $x+1$ can be calculated by the product of their limits.

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

According to the commutative property of multiplication, the above product of two functions can be written as follows.

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

Look at the first factor of the mathematical expression. The limit of $1$ by $x$ plus $1$ can be evaluated by the direct substitution as the value of $x$ approaches to $1$.

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

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

The second factor is similar to the trigonometric limit rule in sine function. So, let’s try to convert the limit of sine of $x$ minus $1$ by $x$ minus $1$ as $x$ approaches to $1$ into trigonometric limit rule in sine function.

If $x\,\to\,1$, then $x-1\,\to\,1-1$. Therefore, $x-1\,\to\,0$

It clears that the value of $x$ minus $1$ tends to zero when the value of $x$ is closer to $1$.

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

Assume $y \,=\, x-1$ for our convenience. Now, transform the function in the second factor position in terms of $y$.

$=\,\,\,$ $\dfrac{1}{2}$ $\times$ $\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{(y)}}{y}}$

$=\,\,\,$ $\dfrac{1}{2}$ $\times$ $\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{y}}{y}}$

According to the trigonometric limit rule in sine function, the limit of sine of angle $y$ by $y$ as the value of $y$ tends to $0$ is equal to one.

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

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

Latest Math Topics

Mar 21, 2023

Feb 25, 2023

Feb 17, 2023

Feb 10, 2023

Jan 15, 2023

Latest Math Problems

Mar 03, 2023

Mar 01, 2023

Feb 27, 2023

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

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

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved