1. ## Limit

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

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

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

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

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

Can you finish?

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

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

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

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

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