Question

How do you find the limit of $\left( 3x+\dfrac{1}{x} \right)-\left( \dfrac{1}{\sin x} \right)$ as ‘x’ approaches 0 using the l’hospital rule?

Answer 1

In the given question, we have been asked to find the limit of a given expression and it is given that we have to find or solve the question using l’hospital rule. In order to solve the question, first we need to differentiate the numerator and denominator with respect to ‘x’ one by one. Then by taking the limit x = 0 as it is given in the question we will put x = 0 in the differential expression. Then simplified further to get the answer.

Complete step by step solution:
We have given that;
$\left( 3x+\dfrac{1}{x} \right)-\left( \dfrac{1}{\sin x} \right)$
Simplifying the above expression by taking LCM, we obtained
LCM of sin(x) and (x) is $x\sin x$
$\dfrac{3{{x}^{2}}\sin x+\sin x-x}{x\sin x}$
Now let,
$f\left( x \right)=3{{x}^{2}}\sin x+\sin x-x$
$g\left( x \right)=x\sin x$
Now,
Differentiating f(x) with respect to ‘x’, we get
$f\left( x \right)=3{{x}^{2}}\sin x+\sin x-x$
$f'\left( x \right)=\dfrac{d}{dx}\left( 3{{x}^{2}}\sin x+\sin x-x \right)$
$f'\left( x \right)=\left( 6x\sin x+3{{x}^{2}}\cos x+\cos x-1 \right)$
Differentiating g(x) with respect to ‘x’, we get
$g\left( x \right)=x\sin x$
$g'\left( x \right)=\dfrac{d}{dx}\left( x\sin x \right)$
$g'\left( x \right)=\sin x+x\cos x$
Now, combining both we will obtained,
$\dfrac{f'x}{g'x}=\dfrac{\left( 6x\sin x+3{{x}^{2}}\cos x+\cos x-1 \right)}{\sin x+x\cos x}$
Taking the limit as ‘x’ approaches to 0, we obtained,
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{f'x}{g'x}=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\left( 6x\sin x+3{{x}^{2}}\cos x+\cos x-1 \right)}{\sin x+x\cos x}=\dfrac{0}{0}$
As we can that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{f'x}{g'x}$ exists or not as we will get the answer in the form $\dfrac{0}{0}$ .

Note: While solving these types of problems, students need to be very careful while doing the calculation part to avoid making any type of error. They need to know about the concept of the finding the value of limit of a given expression using l’hospital rule. Instead of using l’hospital rule we can factorize the polynomial and then put the given value of x i.e. the limit and simplify further. Both the ways we will get the same answer.