<s>[INST] How do i find this limit WITHOUT L'hôpitals rule How do I find 
$$\lim_{x\to 0}\frac{(x+4)^{3/2}+e^x-9}{x}$$
without l'hôpital rule? I know from l'hôpital the answer is 4.
 [/INST] Less elegant than zhw'x answer.
For $(x+4)^{3/2}$, use the generalized binomial theorem of Taylor expansion
$$(x+4)^{3/2}=8+3 x+\frac{3 x^2}{16}+O\left(x^3\right)$$ Using the standard Taylor expnasion of $e^x$, we then have $$(x+4)^{3/2}+e^x-9=\left(8+3 x+\frac{3 x^2}{16}+O\left(x^3\right) \right)+\left( 1+x+\frac{x^2}{2}+O\left(x^3\right)\right)-9$$ $$(x+4)^{3/2}+e^x-9=4 x+\frac{11 }{16}x^2+O\left(x^3\right)$$
$$\frac{(x+4)^{3/2}+e^x-9}{x}=4 +\frac{11 }{16}x+O\left(x^2\right)$$ which shows the limit and how it is approached.
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