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# Using L’ hospital rule evaluate: $\mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^2}\sin x}}$.

Last updated date: 19th Sep 2024
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Hint: According to given in the question we have to evaluate $\mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^2}\sin x}}$ using l’ hospital rule. So, first of all we have to understand the l’ hospital rule as explained below:
L’ hospital Rule: L’ hospital rule can help us to evaluate a limit that can be hard or impossible to solve. According to this rule the limit when we divide one function by another is the same after we take the derivation of each of the functions in the given limit.

Formula used: $\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{{f'}(x)}}{{{g'}(x)}}..................(A)$
Hence, with the help of the formula (A) above, we can evaluate the given limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^2}\sin x}}$
But first of all we have to simplify the given limit $\mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^2}\sin x}}$by multiplying and dividing with x in the numerator and denominator and after that we have to apply the L’ hospital rule to evaluate it.
Formula used:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1.....................(B)$
$\Rightarrow \dfrac{{d\sin x}}{{dx}} = \cos x....................(C) \\ \Rightarrow \dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}................(D) \\ \Rightarrow \dfrac{{d\cos x}}{{dx}} = - \sin x...........(E)$

Complete step-by-step solution:
Step 1: First of all we have to simplify the given limit by multiplying it with x in the numerator and denominator to simplify it. Hence,
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^2}\sin x}} \times \dfrac{x}{x} \\ = \mathop {\lim }\limits_{x \to 0} \dfrac{x}{{\sin x}} \times \dfrac{{x - \sin x}}{{{x^3}}}..........(1)$
Step 2: Now, to solve the obtained expression (1) as in the step 1 we have to apply formula (B) as mentioned in the solution hint.
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^3}}} \times 1 \\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^3}}}................(2)$
Step 2: Now, on applying l’ hospital rule as in the expression (2) as obtained in the solution step 2.
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^3}}}............\dfrac{0}{0}$ form
Now, on applying formula (C) and formula (D) as mentioned in the solution hint,
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{1 - \cos x}}{{3{x^2}}}....................(3)$
Step 3: Now, we have to apply l’ hospital rule again to evaluate the limit in the expression (3) as obtained in the solution step 2. Hence, with the help of the formula (E) as mentioned in the solution hint,
$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{{6x}}$
Step 4: Now, we have to use the formula (B) again to solve the expression as obtained in the solution hint.
$= \mathop {\dfrac{1}{6}\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} \\ = \dfrac{1}{6} \times 1 \\ = \dfrac{1}{6}$

Hence, we have evaluated the given limit with the help of the formulas (A), (B), (C), (D), and (E) as mentioned in the solution hint which is $\mathop {\lim }\limits_{x \to 0} \dfrac{{x - \sin x}}{{{x^2}\sin x}} = \dfrac{1}{6}$

Note: L’ hospital rule allows us to simplify the evaluation of limits that involves intermediate forms and an intermediate form is defined as a limit that does not give enough information to determine that original limit.
L’ hospital rule is the defined way to simplify evaluation of limits. With this we do not evaluate the given limit but first we simplify the given terms of that limit and repeat the process till we finally get the most simplified form.