I have some limit problems and I'm not sure how to do it. I already have the answers so I'm not asking for answers but the "how to".

 

lim (1/x)-(1/3)

-----------

x--->3 (x-3)

 

 

 

lim (t+4)^(1/2)-2

t-->0 -----------

t

 

I always end up getting 0/0 on all the fractions. I'm not sure what to do for these. Thanks!

[math]\frac{0}{0}[/math] is an indeterminate form, which allows you to L'Hopital. So I would suggest you try and apply L'Hopital to these.

[math]\frac{0}{0}[/math] is an indeterminate form, which allows you to L'Hopital. So I would suggest you try and apply L'Hopital to these.

 

I don't think we used L'Hopital or something. Is their a different way?

Okay, just to be clear before I try to help: are these your problems?

 

[math]\lim_{x \to 3} \frac{(1/x) - (1/3)}{x-3}[/math]

 

[math]\lim_{t \to 0} \frac{(t+4)^{\frac{1}{2}} -2}{t}[/math]

 

Sometimes it's hard to tell with equations written in text. You can click on the above formulas to see how I wrote them.

You have it right. Thans for the info!

Well, here's how I'd approach the first one if L'Hopital isn't an option.

 

[math]\lim_{x \to 3} \frac{(1/x) - (1/3)}{x-3} = \lim_{x \to 3}\left( \frac{(1/x)}{x-3} - \frac{(1/3)}{x-3}\right) = \lim_{x \to 3} \left( \frac{1}{x(x-3)} - \frac{1}{3x - 9}\right)[/math]

 

Once you simplify the fractions like that, you can tackle them much more easily.

For the second since L'Hopital is not an option try rationalizing the nuemerator; like this:

 

[math]\frac{(t+4)^{\frac{1}{2}} -2}{t} = \frac{\left(\sqrt{(t+4)} -2\right)\left(\sqrt{(t+4)} +2\right)}{t\left(\sqrt{(t+4)} +2\right)} [/math]

 

[math]\frac{\left(\sqrt{(t+4)} -2\right)\left(\sqrt{(t+4)} +2\right)}{t\left(\sqrt{(t+4)} +2\right)}=\frac{\left(t+4-4\right)}{t\left(\sqrt{t+4} +2\right)}[/math]

 

[math]\frac{\left(t+4-4\right)}{t\left(\sqrt{t+4} +2\right)}=\frac{1}{\sqrt{t+4}+2}[/math]

 

So now we can evaluate the limit and it should be fairly obvious:

 

[math]\lim_{t \to 0} \frac{1}{\sqrt{t+4}+2}=???[/math]