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

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved