How can I solve this limit? (1) lim x---->infinity (sin x) / x (2) lim x---->infinity ( cos x)/x

2 Answers
Jun 27, 2018



Both of these limits are undefined as they alternate from -1, 1 on the top and are unable to converge at #oo#

Jun 27, 2018

Both limits tend to zero.


Both limits are solved in the same way, so I'll explain the general concept with the first one. Everything will work in the exact same way in the second case.

Studying a limit means to understand the behaviour of a function as the input approaches some quantity - in this case, we're asking what happens to the function as the input #x# grows larger and larger.

Since the function is a ratio, let's see what happens to the numerator and denominator. The numerator is the sine function, which waves between #-1# and #1#, and thus does not have a limit as #x\to\infty#. You can visualize it in this way: if you keep going round and round the unit circle, where will you end? Well..nowhere! You just keep going around the circle, without reaching a specific point.

On the other hand, the denominator is #x# itself, and we are assuming that it is growing indefinitely.

So, as long as #sin(x)# does not have a limit as #x\to \infty#, we are sure that #-1 \le sin(x) \le 1#, or #|sin(x)|\le 1#.

This means that we are dividing a quantity that will never exceed one (in absolute value) by a quantity that grows larger and larger. The metaphore of the cake and slices may work fine here: assume you have a cake, and you have to cut it into more and more slice. The greater the number of slice, the smaller each slice will be, wouldn't it?

So, as the numer of slice grows to infinity, there will be almost zero cake in each slice!

To solve this more precisely, we may write

#\lim_{x\to\infty}|\frac{sin(x)}{x}| = \lim_{x\to\infty}\frac{|sin(x)|}{x} \le \lim_{x\to\infty}\frac{1}{x} = 0#

and since #|f(x)|\to 0#, we have #f(x)\to 0#