# 2 Limit Questions

• Jan 19th 2009, 04:43 PM
justinwager
2 Limit Questions
1) lim 1/x+4 - 1/4
------------
x
x->0

This one i have no idea how to start it.....i'm totally lost...

2) lim ------
/ t + 9 - 3
------------- that is suppose to be a square root sign over the t + 9
t

t->0

For this question i got the answer of 1/6 but I dont know how to show my work using this.... forum but if someone can confirm 1/6 is the correct answer...also help me with number 1 thanks!
• Jan 19th 2009, 05:09 PM
mr fantastic
Quote:

Originally Posted by justinwager
1) lim 1/x+4 - 1/4
------------
x
x->0

This one i have no idea how to start it.....i'm totally lost...

2) lim ------
/ t + 9 - 3
------------- that is suppose to be a square root sign over the t + 9
t

t->0

For this question i got the answer of 1/6 but I dont know how to show my work using this.... forum but if someone can confirm 1/6 is the correct answer...also help me with number 1 thanks!

In both cases you can use the theorem that $\displaystyle f'(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}$.

Otherwise you have to hack through the algebra the long way.

• Jan 19th 2009, 05:12 PM
mr fantastic
Quote:

Originally Posted by justinwager
1) lim 1/x+4 - 1/4
------------
x
x->0

This one i have no idea how to start it.....i'm totally lost...

[snip]

As an alternative to a calculus approach you should note that by getting a common denominator you have

$\displaystyle \frac{\frac{1}{x+4} - \frac{1}{4}}{x} = \frac{4 - (x + 4)}{4(x+4)x} = \, ....$
• Jan 19th 2009, 08:11 PM
justinwager
thanks for the help!

based on your help i got

-1/16 on the first question...is that correct?
• Jan 19th 2009, 08:37 PM
mr fantastic
Quote:

Originally Posted by justinwager
thanks for the help!

based on your help i got

-1/16 on the first question...is that correct?

Yes.