Limit of an exponential function

Question

So I have this limit and I was thinking of using L'Hospital's rule. Unfortunately, I am not able to format the equation properly so any help would be great.

$$\lim_{x\to3}e^{\tfrac{x^3-27}{(x-3)(x^3+4)}}$$

First I compute by substitution, and I get an indeterminate form of $e$ raised to the power or $0$ over $0$. So I find the derivative of the power, as to get $\dfrac{3x^2}{3x^2}$. I then substitute $3$ in that and get $e$ to the power of $\dfrac{27}{27}$ which is equal to $e$ to the power of $1$.

Am I correct or is there something missing?

Answer 1

This all comes down to continuity. For any continuous function $f$, $$\lim_{x \rightarrow a} f(g(x)) = f(\lim_{x \rightarrow a} g(x)).$$ So you only need to take the limit of the inside and plug that into $e^x.$ If you factor $x^3-27$ you get $(x - 3) (x^2 + 3 x + 9)$ so you want to take $$\lim_{x\rightarrow 3} \frac{(x - 3) (x^2 + 3 x + 9)}{(x-3)(x^3+4)} = \lim_{x\rightarrow 3} \frac{(x^2 + 3 x + 9)}{(x^3+4)}.$$ Also, for any function continuous at $a$ we have $$ \lim_{x \rightarrow a} f(x) = f(a).$$ You can now compute that limit by plugging in, because the function $\frac{(x^2 + 3 x + 9)}{(x^3+4)}$ is continuous on its domain as well.

Answer 2

No need to use L'hospitals rule

We want to find $$a=\lim_{x \to 3} \frac{x^3-27}{(x-3)(x^3+4)}$$ On the numerator, we can use factor using a difference of cubes to get $(x-3)(x^2+3x+9)$. After canceling this limit can be evaluated.

Our final answer will then be $e^a$

Answer 3

lim(x->3) [e^((x^3-27)/(x-3)(x^3+4)]

lim(x->3) e^(((x-3)(x^2+3x+9)/(x-3)(x^3+4)]

lim(x->3) e^(x^2+3x+9)/(x^3+4)]

= e^(27/24).

The function is not defined at x=3, which of course does not effect the limit.