Question

Can someone explain me why the limit of (sqrt(h^2+4h+5)-5)/H as H approaches 0^+ is equal to (2sqrt5)/5

Answer 1

equal to\[\frac{ 2\sqrt{5} }{ 2 }\]

Answer 2

\[\lim_{h \rightarrow 0^+}\frac{ \sqrt{h^2+4h+5}-\sqrt{5} }{ h }\]

Answer 3

wio are you gonna help me :D?

Answer 4

You can solve it using L'Hopital's rule,
\[\lim_{h\rightarrow0^+}\frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}=\frac{2h+4}{2\sqrt{h^2+4h+5}}=\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}\]

Answer 5

amm I'm not getting \[\frac{ 2h+4 }{ 2\sqrt{h^2+4h+5}}\] where the hell the sqrt(5) goes?

Answer 6

Take the derivative of the numerator,
\[(\sqrt{h^2+4h+5})'=\frac{2h+4}{2\sqrt{h^2+4h+5}}\]
In the last part, you can do,
\[\frac{4}{2\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{\sqrt{5}\sqrt{5}}=\frac{2\sqrt{5}}{5}\]

Answer 7

another way is w/o using L'hop rule :-

\(\large\lim_{h \rightarrow 0^+}\frac{ \sqrt{h^2+4h+5}-\sqrt{5} }{ h } \)

rationalize the numerator

\(\large\lim_{h \rightarrow 0^+}\frac{ \sqrt{h^2+4h+5}-\sqrt{5} }{ h } \times \frac{\sqrt{h^2+4h+5}+\sqrt{5}}{\sqrt{h^2+4h+5}+\sqrt{5}} \)

\(\large\lim_{h \rightarrow 0^+}\frac{ h^2+4h }{ h \times \sqrt{h^2+4h+5}+\sqrt{5}} \)

\(\large\lim_{h \rightarrow 0^+}\frac{ h+4 }{ \sqrt{h^2+4h+5}+\sqrt{5}} \)

\(\large\frac{ 4 }{ \sqrt{5}+\sqrt{5}} \)

\(\large \frac{ 2 }{ \sqrt{5} }\)

\(\large \frac{ 2\sqrt{5}}{5} \)

Answer 8

im getting it know. good way to solve it. My TI-89 is not getting the product of the first step you made. it sucks:(

Answer 9

ohk... not so familiar wid TI-89..