Answer

Verified

421.5k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

A group of fish is known as class 7 english CBSE

The highest dam in India is A Bhakra dam B Tehri dam class 10 social science CBSE

Write all prime numbers between 80 and 100 class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Onam is the main festival of which state A Karnataka class 7 social science CBSE

Who administers the oath of office to the President class 10 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Kolkata port is situated on the banks of river A Ganga class 9 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE