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.
 A: Hint: The expression equals
$$ \frac{f(x) - f(0) + g(x) - g(0)}{x-0}$$
if you choose $f,g$ appropriately.
A: The limit can be found using neither Taylor expansions, nor derivatives, but with much more basic approaches.
First, a consequence of the well-known limit
$$ \lim_{x\rightarrow0} \left(1 +\frac{1}{x}\right)^x = e$$
is
$$ \lim_{x\rightarrow0} \frac{e^x -1}{x} = 1 $$
so the limit in OP is equal to
$$ 1 + \lim_{x\rightarrow0} \frac{(x+4)^{3/2} -8}{x} $$
so long as the second term exists. We calculate the second term as
$$ \lim_{x\rightarrow0} \frac{((x+4)^{3/2}-8)((x+4)^{3/2}+8)}{x((x+4)^{3/2}+8)} = \lim_{x\rightarrow0} \frac{(x+4)^3-64}{x((x+4)^{3/2}+8)} = \\ \lim_{x\rightarrow0} \frac{(x+4)^3-64}{x} \lim_{x\rightarrow0} \frac{1}{((x+4)^{3/2}+8)} = \frac{1}{16} \lim_{x\rightarrow0}\frac{48x+12x^2+x^3}{x}=\frac{48}{16}=3$$
where again then transition from the first line to the second is valid so long as both limits exist (which, as you see, they do).
A: 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.
