Expert: Alon Mandes - 5/19/2010

Question
QUESTION: On my calculus final I had to take a limit for [cos(x) / abs(x)] as x approaches zero. I thought the answer was infinity, because x gets smaller and smaller and cos(0) =1. But there was no answer choice that said infinity. After the exam, I asked my teachers and he said since there is an absolute value at the bottom, the limit doesn't exist. I still don't believe him. Because of the absolute value, the function reaches infinity from both the right and left side, that would mean that the limit does exist.

I even used an online limit calculator:
http://www.numberempire.com/limitcalculator.php

Here is the output:
http://www.numberempire.com/cgi-bin/render2.cgi?\nocache%20\LARGE%20\lim_{x\to0}

The answer did come upto infinity.

My teacher said when you take the l'Hôpital's rule, the answer would be undefined. But I still have a doubt.
Please help, Thank You

ANSWER: Yes, the answer is INFINITY .
You may use a numerical application to sketch the graph. Or simply use Microsoft Excel .


Alon.

---------- FOLLOW-UP ----------

QUESTION: Thank you for you reply.
Is there any way to show that the answer is infinity without graphing?

I tried to use the l'Hôpital's rule.

I separated the function piecewise.

lim(as x approaches 0 from the left)
cos(x) / -x

lim(as x approaches 0 from the right)
cos(x)/ x

Now I take the derivative of the bottom and the top for both pieces.
lim(as x approaches 0 from the left)
-sin(x) / -1 = 0

lim(as x approaches 0 from the right)
-sin(x)/ 1 = 0

What did I do wrong?

Answer
1st of all l'Hôpital's rule can be applied only on form such as INF/INF or 0/0 . In our case
we have 1/0 which doesn't match the category.
The only analytical way is to calculate side limits & show that it's equal :

Lim    Cos(x)/|x| =
x->0+
Lim    Cos(x)/(x) = 1/(0+e) --> INF
x->0+


Lim   Cos(x)/|x| =
x->0-  
Lim   Cos(-x)/(-x) =
x->0-
Lim   -Cos(x)/(x) = -1/(0-e) --> INF
x->0-

Where e "epsilon" is very small positive number .

Alon.