# Evaluating Limits Mathematically

So here's an example of the kind of problem i'm talking about: limits. How exactly do I go about evaluating something like that. I can see and kind of figure it out and the answers make sense, but how do I do this mathematically? Thanks

-

If you look at the fractional exponent $\frac {3}{3-x}$ you can see where you're going. You're approaching 3 from the right side, so the numbers you're going through are larger than 3. For example, 4, 5, 6, etc. So if you were to plug in these values you would get a negative fraction, and $e^{-number}$ =$\frac{1}{e^{number}}$, so you can see that as you get closer to 3 (from the right direction), the $e^{\frac{3}{3-x}}$ gets smaller, i.e. approaches zero.

-

Mathematically: $\lim_{x \to a} f(x) = L$ means: $$\forall \epsilon > 0 \exists \delta > 0$$ so that $$|x - a| < \delta \Rightarrow |f(x) - L| < \epsilon$$

-

so you are trying to figure out $\lim_{x\rightarrow 3^+}\exp(3/(3-x))$ and $\lim_{x\rightarrow 3^-}\exp(3/(3-x))$.

$exp(z)$ is a continuous function so when taking limits, they will pass inside. So $\lim_{x\rightarrow 3^+}\exp(3/(3-x))=\exp(\lim_{x\rightarrow 3^+} 3/(3-x))$ and similarly from the other side. Now, $\lim_{x\rightarrow 3^+}3/(3-x)=\infty$ and $\lim_{x\rightarrow 3^-}3/(3-x)=-\infty$, so it amounts to figuring out $\exp(+\infty)$ and $\exp(-\infty)$ one of which is infinity and the other 0, respectively.

-