Limit evaluation: very tough question, cannot use L'hopitals rule I found a very tough limits question online. The question asks you to evaluate the limit $$\lim_{x \to 0}\frac{(x+4)^\frac{3}{2}+e^{x}-9}{x}$$ without using L'H&ocirc;pitals rule.

I tried to treat the top as a radical expression with the $e^x-9$ grouped and the other in root form to try to attempt rationalization. It did not work because you still get $\frac 0 0$.

I tried a trick of double rationalization but that did not work, got back to the starting.
Second attempt I tried to let $x=z-4$, a substitution, but it still did not lead to something that could remove a zero from the numerator.

Then I tried to break this up into three fractions, by dividing $x$ into each term in the numerator, and I basically got $+\infty$, then can't do $e^x/x$ and then $-\infty$.

So I have exhausted all the algebraic tricks I can think of.

Anybody out there think they they can crack this one?
Hope someone can.

Sincerely,

Palu
Let $f(x) = (x+4)^\frac{3}{2}+e^{_{x}}-9$. Your limit can be written as 
$$\lim_{x \to 0}\frac{f(x)- f(0)}{x - 0}$$
Which is the definition of $f'(0)$. Thus the answer is ${3 \over 2}(0 + 4)^{1 \over 2} + e^0 = 4$.
(x+4)^(3/2)-8/x+e^x-1/x
(x+4)^(3/2)-8/x +1  (standard limit)
((x+4)^(3/2)-4^(3/2))((x+4)^(3/2)+4^(3/2))/x*((x+4)^(3/2)+4^(3/2)) +1
(x+4)^3-4^3/x*((x+4)^(3/2)+4^(3/2)) +1
put x=0
12*4/2*8 +1
4