Question

How do you find the limit of $\dfrac{x-\cos x}{x}$ as x approaches to $\infty$ ?

Answer 1

Hint : In the given question, we have been asked to find the limit of a given function. In order to solve the question first we need to rewrite the expression by expanding the limits because the limit of a sum is the sum of the limits and then we have to solve the first term given in the expression using L-hospital rule. And the second term by squeeze theorem. Then combined the both terms and simplified further to get the answer.

Complete step-by-step answer :
We have given that,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{x-\cos x}{x} \right)$
Rewritten the above as,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{x-\cos x}{x} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{x}{x}-\dfrac{\cos x}{x} \right)$
Separating the terms, we obtained
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{x}{x} \right)-\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\cos x}{x} \right)$
As we know that,
Using the L-hospital’s rule i.e. $\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$
$\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{x}{x} \right)=\dfrac{1}{1}=1$
Now, we have
$\Rightarrow 1-\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\cos x}{x} \right)$
Now,
Taking $\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\cos x}{x} \right)$
As we know that the range of the trigonometric function $\cos x$ is $-1\le \cos x\le 1$
Using the squeeze theorem that has three steps to follow i.e. if $g\left( x \right)\le f\left( x \right)\le h\left( x \right)\$ and $\underset{x\to \infty }{\mathop{\lim }}\,g\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,h\left( x \right)\ =L$ , then $\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)\ =L$ .
Applying the theorem, we obtained
Let $f\left( x \right)\ =\dfrac{1}{x}\cos x,\ g\left( x \right)=-\dfrac{1}{x}\ and\ h\left( x \right)\ =\dfrac{1}{x}$
Thus,
$\Rightarrow -\dfrac{1}{x}\ \le \dfrac{1}{x}\cos x\le \dfrac{1}{x}$
Now,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{x} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( -\dfrac{1}{x} \right)=0=L$
As we are dividing a very small number i.e. 1 by a large number that is continually increasing.
Therefore,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{x}\cos x \right)=\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\cos x}{x} \right)=L=0$
Thus, at last
We have,
$\Rightarrow 1-\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{\cos x}{x} \right)=1-0=1$
Therefore,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{x-\cos x}{x} \right)=1$
Hence, this is the required answer.

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 which states that $\underset{x\to \infty }{\mathop{\lim }}\,\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$ and by using the squeeze theorem which states that if $g\left( x \right)\le f\left( x \right)\le h\left( x \right)\$ and $\underset{x\to \infty }{\mathop{\lim }}\,g\left( x \right)=\underset{x\to \infty }{\mathop{\lim }}\,h\left( x \right)\ =L$ , then $\underset{x\to \infty }{\mathop{\lim }}\,f\left( x \right)\ =L$ .