Question

What is the limit of $\left( x-\dfrac{\cos x}{x} \right)$ as x goes to infinity?

Answer 1

Assume the required limit as L. First of all break the terms of this of the limit and write it as the difference of two limits given as $\displaystyle \lim_{x \to \infty }\left( x \right)-\displaystyle \lim_{x \to \infty }\left( \dfrac{\cos x}{x} \right)$. Now, check the values of the functions as x tends to infinity. Use the fact that the range of the cosine function is $\left[ -1,1 \right]$ to evaluate the second part of the relation.

Complete step by step answer:
Here we have been asked to find the limit of the function $\left( x-\dfrac{\cos x}{x} \right)$ as the domain value, i.e. x, tends to infinity. Let us assume the limit value as L so mathematically we have,
$\Rightarrow L=\displaystyle \lim_{x \to \infty }\left( x-\dfrac{\cos x}{x} \right)$
Now, we can consider the given function as the difference of two functions and we know that $\displaystyle \lim_{x \to a}\left( f\left( x \right)\pm g\left( x \right) \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)\pm \displaystyle \lim_{x \to a}g\left( x \right)$, so we can write the above expression of the limit as:
$\Rightarrow L=\displaystyle \lim_{x \to \infty }\left( x \right)-\displaystyle \lim_{x \to \infty }\left( \dfrac{\cos x}{x} \right)$
We know that the value of the cosine function oscillates from -1 to 1 so its range is $\left[ -1,1 \right]$ irrespective of the value of its argument (angle). So, we can say that as x tends to infinity the ratio $\dfrac{\cos x}{x}$ tends to 0 because the denominator will become very large and the numerator will be very small. Therefore we get,
$\begin{aligned} & \Rightarrow L=\displaystyle \lim_{x \to \infty }\left( x \right)-0 \\\ & \Rightarrow L=\displaystyle \lim_{x \to \infty }\left( x \right) \\\ \end{aligned}$
Now, as x will tend to infinity its limit value will also tend to infinity because y = x is linear function and the value of y increases will increase in the value of x. So we get,
$\Rightarrow L=\infty$
We know that infinity is not real and it is also not finite, hence we can conclude that the limit of the given function approaches infinity.

Note: Remember the formulas $\displaystyle \lim_{x \to a}\left( f\left( x \right)\pm g\left( x \right) \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)\pm \displaystyle \lim_{x \to a}g\left( x \right)$, $\displaystyle \lim_{x \to a}\left( f\left( x \right)\times g\left( x \right) \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)\times \displaystyle \lim_{x \to a}g\left( x \right)$ and $\displaystyle \lim_{x \to a}\left( f\left( x \right)\div g\left( x \right) \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right) \right)\div \displaystyle \lim_{x \to a}g\left( x \right)$. Here we cannot apply any formula to get the answer but we need to understand the condition geometrically. We cannot use the L Hospital’s rule as the function is not of the form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$. We can convert it into the form $\dfrac{\infty }{\infty }$ by taking the L.C.M but after using the rule we will get $\displaystyle \lim_{x \to \infty }\left( 2x-\sin x \right)$ which will again give the answer $L=\infty$.