Question

Need help on two problems, 23 and 25. Can anyone help me out?

Answer 1

for 25 ,
rationalize the numerator

Answer 2

For 23 I keep getting the same answer. 1/9 but it's supposed to be -1/9
And for 25 I lost completely

Answer 3

I'm not sure how to do that with square roots

Answer 4

23.\[\lim_{x \rightarrow 3}\frac{ \frac{ 1 }{ x }-\frac{ 1 }{ 3 } }{ x-3 }=\lim_{x \rightarrow 3}\frac{ \frac{ 3-x }{ 3x } }{ x-3 }\]
\[=-\lim_{x \rightarrow 3}\frac{ x-3 }{ 3x(x-3) }=-\lim_{x \rightarrow 3}\frac{ 1 }{ 3x }=-\frac{ 1 }{ 9 }\]

Answer 5

multiply top and bottom by \(\sqrt{1+t}+\sqrt{1-t}\)

Answer 6

25
\[\lim_{t \rightarrow 0}\frac{ \sqrt{1+t} -\sqrt{1-t}}{ t }\times \frac{ \sqrt{1+t}+\sqrt{1-t } }{ \sqrt{1+t}+\sqrt{1-t} }\]
\[=\lim_{t \rightarrow 0}\frac{ \left( \sqrt{1+t} \right)^2-\left( \sqrt{1-t} \right)^2 }{ t(\sqrt{1+t}+\sqrt{1-t}) }\]
\[=\lim_{t \rightarrow 0}\frac{ 1+t-(1-t) }{ t(\sqrt{1+t} +\sqrt{1-t)}}\]
=?

Answer 7

If you, briens, would show your own calculations, it'd be a lot easier for others to give you meaningful feedback.

Answer 8

Here are some similar probs, with the solutions that start below them on pg 4...lots of algebra really to practice and the limit properties

Answer 9

Wow thank you guys so much!!!