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'Hôpitals 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
 A: 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 ${3 \over 2}(0 + 4)^{1 \over 2} + e^0 = 4$.
A: This one is easily transformed into an algebraic limit. We need to make use of the following standard limit theorems $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ We can proceed in the following manner
\begin{align}
L &= \lim_{x \to 0}\frac{(x + 4)^{3/2} + e^{x} - 9}{x}\\
&= \lim_{x \to 0}\frac{(x + 4)^{3/2} - 8}{x} + \frac{e^{x} - 1}{x}\\
&= \lim_{x \to 0}\frac{(x + 4)^{3/2} - 8}{(x + 4) - 4} + 1\\
&= \lim_{t \to 4}\frac{t^{3/2} - 4^{3/2}}{t - 4} + 1\text{ (by putting }t = x + 4)\\
&= \frac{3}{2}\cdot 4^{1/2} + 1 = 4
\end{align} 
A: Rewrite (x+4)^(3/2) as [4^(3/2) (1+x/4)^(3/2)] which is 8 (1+x/4)^(3/2). When x is small, (1+x/4)^(3/2) can be approximated by [1+(3/2)(x/4)] that is to say [1 + 3 x / 8]; then, close to zero, (x+4)^(3/2) is (8 + 3 x). On the other hand, when x is close to zero, Exp[x] can be approximated by (1 + x). Then, the numerator is approximated by : 8 + 3 x + 1 + x - 9 = 4 x. So the result is (4 x / x) = 4.
A: f(x)=(〖(x+4)〗^(3/2)+e^x-9)/x
According to Maclauren's Series
f(x)=(x^0 f(0))/0!+(x^1 f^' (0))/1!+(x^2 f^'' (0))/2!+(x^3 f^''' (0))/3!+⋯
So, using this series
We get
1)   〖(x+4)〗^(3/2)=8+3x+3/16 x^2+⋯
2) e^x=1+x+x^2/2+⋯
Substituting these in f(x)
We'll get
f(x)=((8+3x+3/16 x^2+⋯)+(1+x+x^2/2+⋯)-9)/x
Combining terms of same powers we get
f(x)=((8+1-9)+(3+1)x+(3/16+1/2) x^2)/x
Therefore, 
f(x)=4+8/16 x+⋯
The rest of the terms have higher degrees of x
So, now applying limits, we get 
lim┬(x→0)⁡〖f(x)=lim┬(x→0)⁡〖(4+8/16 x+⋯〗 〗)=4
A: Just take 4 comman and use binomial expansion and then expand e^x and solve very simple
Not much calculation
A: (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
