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I'm pretty sure that this uses L'Hopital's rule but when I try to apply this I end up going in circles and am unable to ever find a limit.

Taking derivatives of numerator and denominator:

$\lim_{h \to 0} \frac{-sin(x+h) +sin(x)}{\frac{1}{2}({x+h})^{-1/2}-\frac{1}{2}{x}^{-1/2}}$

This still results in $\frac{0}{0}$ though.

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  • Do you understand that the limit you wrote in the title is different from the one you've got in the post? – Michael Rybkin Nov 15 '17 at 06:04
  • You didn't use L'Hopital's rule correctly. You should have differentiated the numerator and denominator with respect to $h$, not to $x$. – symplectomorphic Nov 15 '17 at 07:55
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    Sorry but you should be able to identity, without L'H of course, the limits $$\lim_{h\to0}\frac{\cos(x+h)-\cos(x)}h$$ and $$\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}h$$ simply as derivatives of well known functions. And now, $$\frac{\cos(x+h)-\cos(x)}{\sqrt{x+h}-\sqrt{x}}=\frac{\frac{\cos(x+h)-\cos(x)}h}{\frac{\sqrt{x+h}-\sqrt{x}}h}$$ hence you are done. To sum up: Forget L'H. – Did Nov 15 '17 at 09:10

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$\lim_{h\rightarrow 0}\dfrac{\cos(x+h)-\cos x}{\sqrt{x+h}-\sqrt{x}}=\lim_{h\rightarrow 0}\dfrac{\cos(x+h)-\cos x}{h}\cdot(\sqrt{x+h}+\sqrt{x})=-2\sqrt{x}\sin x$.

user284331
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