anonymous
  • anonymous
Can someone explain me why the limit of (sqrt(h^2+4h+5)-5)/H as H approaches 0^+ is equal to (2sqrt5)/5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
equal to\[\frac{ 2\sqrt{5} }{ 2 }\]
anonymous
  • anonymous
\[\lim_{h \rightarrow 0^+}\frac{ \sqrt{h^2+4h+5}-\sqrt{5} }{ h }\]
anonymous
  • anonymous
wio are you gonna help me :D?

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John_ES
  • John_ES
You can solve it using L'Hopital's rule, \[\lim_{h\rightarrow0^+}\frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}=\frac{2h+4}{2\sqrt{h^2+4h+5}}=\frac{4}{2\sqrt{5}}=\frac{2\sqrt{5}}{5}\]
anonymous
  • anonymous
amm I'm not getting \[\frac{ 2h+4 }{ 2\sqrt{h^2+4h+5}}\] where the hell the sqrt(5) goes?
John_ES
  • John_ES
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, \[\frac{4}{2\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{\sqrt{5}\sqrt{5}}=\frac{2\sqrt{5}}{5}\]
ganeshie8
  • ganeshie8
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 \(\large\lim_{h \rightarrow 0^+}\frac{ \sqrt{h^2+4h+5}-\sqrt{5} }{ h } \times \frac{\sqrt{h^2+4h+5}+\sqrt{5}}{\sqrt{h^2+4h+5}+\sqrt{5}} \) \(\large\lim_{h \rightarrow 0^+}\frac{ h^2+4h }{ h \times \sqrt{h^2+4h+5}+\sqrt{5}} \) \(\large\lim_{h \rightarrow 0^+}\frac{ h+4 }{ \sqrt{h^2+4h+5}+\sqrt{5}} \) \(\large\frac{ 4 }{ \sqrt{5}+\sqrt{5}} \) \(\large \frac{ 2 }{ \sqrt{5} }\) \(\large \frac{ 2\sqrt{5}}{5} \)
anonymous
  • anonymous
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:(
ganeshie8
  • ganeshie8
ohk... not so familiar wid TI-89..

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