How would I evaluate $\lim_{x \to \infty}x\int_0^{1/x}e^t \cos(t) \space \text{d}t$? I need to evaluate $$\lim_{x \to \infty}x\int_0^{1/x}e^t \cos(t) \space \text{d}t$$ and I'm not really sure how to start. Do I have to find the integral or is there another way to figure it out?

Substituting $u = 1/x$, your limit is the same as
$$\lim_{u\to 0} \frac{\int_0^u e^t\cos t\, dt}{u},$$
which by L'hospital's rule and fundamental theorem of calculus is
$$\lim_{u\to 0}  \frac{e^u \cos u}{1} = 1.$$

The problem is asking for the average value of $f(t)=e^t\cos t$ over the interval $[0,1/x]$.  Since $f(t)$ is continuous everywhere, its average value on a small interval near zero approaches $f(0)$, namely $1$.

The simplest approach is through L'Hospital's Rule and the Fundamental Theorem. But there are more complicated ways.
We examine 
$$\int_0^{1/x}e^t\cos t\,dt$$
for large positive $x$. Note that $e^t\cos t$ is increasing in the interval $(0,\pi/4)$ (look at the derivative). So if $x$ is large, and $0\le t\le 1/x$, we have $e^t\cos t\lt e^{1/x}\cos(1/x)$. It follows that
$$\frac{1}{x}\lt \int_0^{1/x} e^t\cos t\,dt\lt \frac{1}{x}e^{1/x}\cos(1/x),$$
and therefore
$$1\lt x\int_0^{1/x} e^t\cos t\,dt\lt e^{1/x}\cos(1/x).$$
Now Squeeze.