Convergence of a sequence of integrals 
I've tried expanding the hinted expression by using the definition from part (i) and choosing an X0 sufficiently large that |f(x)-l| < 1 but this doesn't appear to help very much at all. I've gotten a bunch of expressions that appear to tend to zero as X goes to infinity but I've hit a brick wall completely. 
 A: $f(x) \to \ell$, so for any $\varepsilon>0$ there is an $X_0>0$ such that $\lvert f(x)-\ell \rvert < \frac{1}{2}\varepsilon $. Then
$$ \frac{1}{X}\int_0^X f(x) \, dx - \ell = \frac{1}{X}\int_0^X (f(x)-\ell) \, dx $$
Now apply the triangle inequality, $\lvert\int\rvert \leqslant \int \lvert \rvert$, and split the integral as the question suggests, so
$$ \left\lvert \frac{1}{X}\int_0^X f(x) \, dx - \ell \right\rvert \leqslant \frac{1}{X}\int_0^{X_0} \lvert f(x)-\ell \rvert \, dx + \frac{1}{X}\int_{X_0}^X \lvert f(x)-\ell \rvert \, dx. $$
Now, $f(x)-\ell$ is continuous, so bounded above, blah blah blah, so the first integral is bounded by a constant times $1/X$, so we can make it as small as we like by choosing $X$ large enough (insert epsilon argument here), and hence smaller than $\frac{1}{2}\varepsilon$. The second integral has an integrand bounded above by $\frac{1}{2}\varepsilon$, so the second integral is bounded above by 
$$\frac{X-X_0}{X}\tfrac{1}{2}\varepsilon = \left( 1-\frac{X_0}{X} \right) \tfrac{1}{2}\varepsilon < \tfrac{1}{2}\varepsilon, $$
and hence the sum of both terms is smaller than $\varepsilon$.
A: $$f\;\;\text{is continuous}\implies\;\text{it has a primitive function, say}\;\;F\implies$$
$$\lim_{X\to\infty}\frac1X\int_0^X f(x)dx=\lim_{X\to\infty}\frac{F(X)-F(0)}X\stackrel{l'Hospital}=\lim_{X\to\infty}F'(X)=\lim_{X\to\infty}f(X)=L$$
