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I need to find the limit to the following:

$$\lim_{h\to0} \frac{\sqrt{1+h}-1}{\tan(h)}$$

and I actually did find it quite easily using l'Hospital's rule.

However we haven't gone through that in the course yet, is there a different approach you can take in solving this problem?

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    Are you sure the numerator shouldn't be $\sqrt{1+h}-1$? – 2017-02-05
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    Yeah, I just noticed that I missed out on the -1. – 2017-02-05
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    @ClementC. Thank you for the foresight – 2017-02-05

2 Answers 2

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Hint:

$$\begin{align}\frac{\sqrt{1+h}-1}{\tan(h)}&=\frac{\sqrt{1+h}-1}h\frac{\sqrt{1+h}+1}{\sqrt{1+h}+1}\frac h{\sin(h)}\cos(h)\\&=\frac1{\sqrt{1+h}+1}\frac h{\sin(h)}\cos(h)\end{align}$$

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    Small additional detail: what you wrote doesn't rule out the limit being $\infty$ (here, it's not even diverging to $\infty$) – 2017-02-05
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    @ClementC. I kind of shrug off limits being $\pm\infty$, but sure, I suppose so. – 2017-02-05
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    Hi, sorry, I completely missed that it's supposed to be sqrt(1+h)-1/tan(h) – 2017-02-05
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    @ejbs No problem, re-updated – 2017-02-05
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    @SimplyBeautifulArt In the middle step I can see that you've used tan(h) = sin(h)/cos(h), did you then also multiply with h/h and then finally multiply with sqrt(1+h)+1/sqrt(1+h)+1 ? Nice, so it can be done with 'normal' re-writes. And then you used (x+y)(x-y) = x^2-y^2 it looks like. Anyway, thank you! – 2017-02-05
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    @ejbs Yup, you caught all of the steps! :D – 2017-02-05
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Equivalents:

$\sqrt{1+h}-1\sim_0\frac12 h$, $\;\tan h\sim_0 h$, so $$\frac{\sqrt{1+h}-1}{\tan h}\sim_0\frac{\frac12 h}{h}=\frac12.$$