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

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 $x^2-1$.

Eliminate the indeterminate form by factoring

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.

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

Find the Limit by Direct Substitution

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 $1$.

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

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

Find the Limit by the Trigonometric Limit rule

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.

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

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

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