# Thread: Can someone help me?

1. ## Can someone help me?

lim [(e^3x)-1]/5x = [(e^0)-1]/5(0)=1-1/0=0/0, so the limit does not exist
x-->0

Is the conclusion that was drawn correct?If,so explain why.If not why not.If you believe the limit in the question exists,then propose a value for the limit and support your answer.

2. Originally Posted by mathcalculushelp
lim [(e^3x)-1]/5x = [(e^0)-1]/5(0)=1-1/0=0/0, so the limit does not exist
x-->0

Is the conclusion that was drawn correct?If,so explain why.If not why not.If you believe the limit in the question exists,then propose a value for the limit and support your answer.
You must be careful with that line of reasoning here is an example.

$\displaystyle \lim_{x \to 1}\frac{x-1}{x^2-1}$

using your logic above when we evaluate at $\displaystyle x=1$ you get

$\displaystyle \lim_{x \to 1}\frac{x-1}{x^2-1}=\frac{1-1}{1-1}=\frac{0}{0}$ is undefinded

but that is not correct now factor the denominator and reduce to get

$\displaystyle \lim_{x \to 1}\frac{x-1}{x^2-1}=\lim_{x \to 1}\frac{x-1}{(x-1)(x+1)}=\lim_{x \to 1}\frac{1}{x+1}=\frac{1}{2}$

To convince your self draw a plot of $\displaystyle f(x)=\frac{x-1}{x^2-1}$

Then graph the function $\displaystyle g(x)=\frac{e^{3x}-1}{5x}$ and guess what the limit will be.

This is called an indeterminate form i.e $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ both types of limits can be calculated using L'Hosiptials rule or better yet infinite series.