Question

lim(h)/[(sqrt(2x+2h-5)) - (sqrt(2x-5))] h->0

Answer 1

Hello and Welcome to OpenStudy! :)
do you mean something like:
\[\lim_{h \rightarrow 0} (\sqrt{2x + 2h-5} - \sqrt{2x-5})\]

Answer 2

yes but with h being a numerator

Answer 3

\[
\lim_{h \rightarrow 0} \frac{h}{(\sqrt{2x + 2h-5} - \sqrt{2x-5})}
\]

Answer 4

right, so first I put the denominator all under one squareroot
so the denominator becomes sqrt(2x-2x-5+5+2h) which you can then add like terms and cancel out some stuff

Answer 5

\[
\lim_{h \rightarrow 0} \frac{h}{(\sqrt{2x + 2h-5} - \sqrt{2x-5})} = \\
\lim_{h \rightarrow 0} \frac{h}{(\sqrt{2x + 2h-5} - \sqrt{2x-5})} *
\frac{(\sqrt{2x + 2h-5} + \sqrt{2x-5})}{(\sqrt{2x + 2h-5} + \sqrt{2x-5})}= ?\\

\]

Answer 6

oh right, that's the way to do it ^_^"

Answer 7

\[
\lim_{h \rightarrow 0} \frac{h}{(\sqrt{2x + 2h-5} - \sqrt{2x-5})} = \\
\lim_{h \rightarrow 0} \frac{h}{(\sqrt{2x + 2h-5} - \sqrt{2x-5})} *
\frac{(\sqrt{2x + 2h-5} + \sqrt{2x-5})}{(\sqrt{2x + 2h-5} + \sqrt{2x-5})}= \\
\lim_{h \rightarrow 0} \frac{h(\sqrt{2x + 2h-5} + \sqrt{2x-5})}{2x+2h-5-2x+5} = \\
\lim_{h \rightarrow 0} \frac{h(\sqrt{2x + 2h-5} + \sqrt{2x-5})}{2h} = \\
\lim_{h\rightarrow 0} \frac{(\sqrt{2x + 2h-5} + \sqrt{2x-5})}{2} = \sqrt{2x-5}\\
\]

Answer 8

Thanks, i didnt see the 2h at the bottom

Answer 9

The h at the bottom cancels the h at the top left.