Can anyone help me to find? the #lim_(x rarr 0) (xcos2x)/(cos5x )#

1 Answer
Oct 20, 2017

The answer is 0.

Explanation:

First, we should know that when #lim_(x->0) f(x)#and #f(x)# is a fraction, the constrain is that the denominator cannot be #0#.

In this case, #lim_(x->0) (xcos2x)/(cos5x)# , if we just sub #x=0# into the denominator, it will become #cos(5*0)=cos0=1#, which is not equal to #0#.

So, we can just simply sub #x=0# in this case and get the following:

#lim_(x->0) (xcos2x)/(cos5x)=(0)(cos2(0))/(cos5(0)] =(0)(1)/(1)=0#