Limit evaluation: very tough question, cannot use L'hopitals rule

Question

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

Answer 1

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 ${3 \over 2}(0 + 4)^{1 \over 2} + e^0 = 4$.

Answer 2

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.