1. ## Limit

Prove that if $f(x) and $\lim_{x \to \infty}f(x) = \lim_{x \to \infty}h(x)=L$, then $\lim_{x \to \infty}g(x)=L$.

Can we prove this by contradiction, assuming to the contrary that $\lim_{x \to \infty}g(x)\not = L$?

2. Here are several observations.
We have $0 \leqslant g(x) - f(x)\, \leqslant h(x) - f(x) \Rightarrow \left| {g(x) - f(x)} \right| \leqslant \left| {h(x) - f(x)} \right|\,$.

We know that if $\varepsilon > 0$ then $\left( {\exists K > 0} \right)\left[ {x \geqslant K\, \Rightarrow \,\left| {f(x) - L}\right| < \frac{\varepsilon }{3}\;\& \,\left| {h(x) - L} \right| < \frac{\varepsilon }{3}} \right]$.

Note that $\left| {g(x) - L} \right| \leqslant \left| {g(x) - f(x)} \right| + \left| {f(x) - L} \right| \leqslant \left| {h(x) - f(x)} \right| + \left| {f(x) - L} \right|$.

Can you finish?

3. You did 99% of it. The rest is easy.

I did it by proving $\lim_{x \to \infty}(h(x)-f(x) )= 0$, then I showed that

$|h(x)-g(x)-(L-L)| =|h(x)-g(x)|\leq |h(x)|-|g(x)|<|h(x)-f(x)-(L-L)|=|h(x)-f(x)| < ?$

Then realized that there wasn't direct connection to $\epsilon$--hmm not too smart.

4. $\dfrac{\varepsilon }
{3} + \dfrac{\varepsilon }
{3} + \dfrac{\varepsilon }
{3} = \varepsilon$