Math Doubts

Evaluate $\displaystyle \lim_{x \to 1} \dfrac{\sin(x-1)}{x^2 -1}$

The angle of a right angled triangle is expressed as $x-1$. The value of the ratio of $\sin(x-1)$ to $x^2 -1$ is required to find when the value of limit $x$ tends to $1$.

Step: 1

Substitute $x = 1$ to test this function.

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

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

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

We have obtained an indeterminate form in substitution method when the value of $x$ is replaced by $1$. So, it has to be solved in another way.

In numerator, the angle is $x-1$ and it can be obtained in denominator as well by expressing $x^2 -1$ as two multiplying factors in which $x-1$ is one of them.

$\implies \Large \displaystyle \lim_{x \to 1} \normalsize \dfrac{\sin(x-1)}{x^2 -1}$ $=$ $\Large \displaystyle \lim_{x \to 1} \normalsize \dfrac{\sin(x-1)}{(x-1)(x+1)}$

Step: 2

Split the ratio of the functions as two multiplying factors.

$= \Large \displaystyle \lim_{x \to 1} \normalsize \Bigg[ \dfrac{\sin(x-1)}{(x-1)} \times \dfrac{1}{(x+1)} \Bigg]$

$= \Large \displaystyle \lim_{x \to 1} \normalsize \Bigg[ \dfrac{\sin(x-1)}{(x-1)}\Bigg] \times \Bigg[\dfrac{1}{(x+1)} \Bigg]$

Step: 3

Apply the limit to both multiplying functions by the product rule.

$= \Large \displaystyle \lim_{x \to 1} \normalsize \Bigg[ \dfrac{\sin(x-1)}{(x-1)} \Bigg]$ $\times$ $\Large \displaystyle \lim_{x \to 1} \normalsize \Bigg[\dfrac{1}{(x+1)} \Bigg]$

Step: 4

The first multiplying factor represents the relation of sine of angle with angle and it is same the limit x tends to 0 sinx/x rule. According to this identity, the quotient of sine of angle by angle is one when $x$ approaches zero. So, let us try to adjust the first multiplying factor.

If $x$ tends to $1$, then $x-1$ tends to $0$. So, change the limit value $x \to 1$ to $x-1 \to 0$ but only for the first multiplying function.

$= \Large \displaystyle \lim_{x-1 \to 0} \normalsize \Bigg[ \dfrac{\sin(x-1)}{(x-1)} \Bigg]$ $\times$ $\Large \displaystyle \lim_{x \to 1} \normalsize \Bigg[\dfrac{1}{(x+1)} \Bigg]$

Therefore, the value of first multiplying factor is $1$ and substitute $x=1$ in the second multiplying factor to obtain the required solution.

$= 1 \times \dfrac{1}{1+1}$

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

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



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