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I am trying to find the limit $$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt{5}}{h}$$

My approach is to divide both the numerator and denominator by $h=\sqrt{h^2}$ since $h>0$. Then we get $$\lim_{h\to0^+}\sqrt{\frac{h^2+4h+5}{h^2}}-\sqrt{\frac{5}{h^2}}=\lim_{h\to0^+}\sqrt{1+\frac{4}{h}+\frac{5}{h^2}}-\sqrt{\frac{5}{h^2}}$$

I am stuck because the limit is when $h\to0^+$ not $h\to\infty$. How can I make it into simpler functions when we already know the limit or when we can just substitute 0 into the function?

Helps are greatly appreciated. Thanks!

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    $\begingroup$ You're on the right track with a clever form of $1$ but you used the wrong clever form of $1$. What you want to do is "rationalize" the numerator by multiplying by $$\frac{\sqrt{h^2+4h+5}+\sqrt{5}}{\sqrt{h^2+4h+5}+\sqrt{5}}.$$ $\endgroup$ – Cameron Williams Jul 29 '14 at 14:17
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$$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt{5}}{h}=\lim_{h\to0^+}\frac{(\sqrt{h^2+4h+5}-\sqrt{5})(\sqrt{h^2+4h+5}+\sqrt{5})}{h(\sqrt{h^2+4h+5}+\sqrt{5})}=\lim_{h\to0^+}\frac{h+4}{(\sqrt{h^2+4h+5}+\sqrt{5})}=\frac2{\sqrt5}$$

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$$\lim_{h\to0^+}\frac{\sqrt{h^2+4h+5}-\sqrt{5}}h=\frac{d(\sqrt{x^2+4x+5})}{dx}_{(\text{ at }x=0)}$$

$$=\frac{2x+4}{2\sqrt{x^2+4x+5}}_{(\text{ at }x=0)}$$

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    $\begingroup$ Damn! I need to type faster :) $\endgroup$ – alandella Jul 29 '14 at 14:29
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    $\begingroup$ @AndreaL. Don't worry! lab is the most faster and if you beat him you realize a new record :) $\endgroup$ – user63181 Jul 29 '14 at 22:15

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