How do i find this limit WITHOUT L'hôpitals rule

Question

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.

Answer 1

Hint: The expression equals

$$ \frac{f(x) - f(0) + g(x) - g(0)}{x-0}$$

if you choose $f,g$ appropriately.

Answer 2

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.

Answer 3

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).