 QUESTION

# Evaluate the value of the following limit: $\mathop {\lim }\limits_{x \to 0} \dfrac{{{\text{x - sinx}}}}{{{{\text{x}}^3}}}$

Hint: To solve this question we will use L’ Hospital rule to evaluate the value of the given limit. Then at last use the direct result of $\dfrac{sinx}{x}$ when the limit of $x$ tends to zero.

Now, the given limit becomes infinite when we put the value of x. So, we will use the L’ Hospital rule to find the value of the limit. The rule is applicable when there is $\dfrac{0}{0}$, $\dfrac{\infty }{\infty }$ form on putting the limit. In the given limit we get the $\dfrac{0}{0}$ . L’ Hospital rule states that if we get the form stated above, we will differentiate both numerator and denominator independently until we get a finite limit.
So, given limit is $\mathop {\lim }\limits_{x \to 0} \dfrac{{{\text{x - sinx}}}}{{{{\text{x}}^3}}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{\text{1 - cosx}}}}{{{\text{3}}{{\text{x}}^2}}}$
Now, putting the value of limit, we can see that there is still a $\dfrac{0}{0}$ form. So, again applying L’ Hospital rule, we get
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin {\text{x}}}}{{6x}}$
$\Rightarrow$ $\dfrac{1}{6}\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin {\text{x}}}}{{\text{x}}}$
Now, as $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin {\text{x}}}}{{\text{x}}}{\text{ = 1}}$
So, $\mathop {\lim }\limits_{x \to 0} \dfrac{{{\text{x - sinx}}}}{{{{\text{x}}^3}}}{\text{ = }}\dfrac{1}{6}$
So, the value of the given limit is $\dfrac{1}{6}$.
Note: When we come up with such types of problems in which we have to find the value of the limit. In such questions, first put the value of the given limit and check whether the value is finite or infinite. If the value comes infinite and of the form of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then the easiest method to find answer is by applying the L’ Hospital rule. You have to apply L’ Hospital rule until the limit becomes finite.