# Math Help - Evaluating limit

1. ## Evaluating limit

Did l work out this problem correctly ? I thought there was a step which l did wrongly. Can you pliz help

BTW the question and attempt to question is on attached image

2. Originally Posted by nyasha
Did l work out this problem correctly ? I thought there was a step which l did wrongly. Can you pliz help

BTW the question and attempt to question is on attached image
You have a mistake in the 2nd line.
How in earth did you get $x^{\frac{3}{2}}$ ?

3. Originally Posted by nyasha
Did l work out this problem correctly ? I thought there was a step which l did wrongly. Can you pliz help

BTW the question and attempt to question is on attached image
You have a mistake in the 2nd line.
How in earth did you get $x^{\frac{3}{2}}$ ?

4. I was wondering why he would write $x^3$ as $\left(x^{3/2}right)^2$ myself but I notice it disappears in the next line!

Nyasha, Yes, what you have done is correct. Although, in fact, it was not necessary to be that general. Just showing that
$\lim_{x\to 0}\frac{x^3}{3x^3}= \frac{1}{3}$
where the limit is taken along the horizontal line y= -2, and
$\lim_{y\to -2}\frac{4(y+ 2)^3}{-(y+2)^2}= 0$
where the limit is taken along the vertical line x= 0, are different is sufficient to show that the limit does not exist.

If you were doing that thinking that if you got a result that did NOT depend on m that would tell you that the limit did exist and would give you the limit, you are mistaken. For one thing, the vertical line, x= constant, cannot be written in the form y= mx so you would not have proven that the limit along all straight lines is the same. And even if you did that, it would not be enough. There exist examples where the limit along any straight line gives the same thing but limits taken along curves gives a different result- in those cases, the limit itself still does not exist.