# Using definition of limits to prove

Given two functions $f, g : A \to \mathbb{R}$ and a cluster point $x_0$ for $A$, define $h(x) = f(x) + 2 g(x)$ and assume $\lim_{x \to x_0} f(x) = L$ and $\lim_{x \to x_0} g(x) = M$.

By applying ONLY the definition of limit, prove that $\lim_{x \to x_0} h(x) = L + 2M$ (Note you are NOT allowed to use algebra of limits)

Okay so I've started this question but I'm sure how to finish it. So far I have:

Given $\varepsilon$>0 and $\delta$>0 there exists $|x-x_0|$< $\delta_1$ implying $|f(x)-L|<\epsilon$ and $|x-x_0|$< $\delta_2$ implying $|g(x)-M|<\epsilon$

Now we have $|f(x)+2g(x)-L-2M| = |f(x)-L|+2|g(x)-M|$

I'm not sure how to manipulate it further to get it in terms of $\delta$.

• Hint : Use $\large \frac{\epsilon}{3}$ as an upper bound for $|f(x)-L|$ and $|g(x)-M|$.Then use the triangular inequality for absolute values. – Peter Aug 13 '15 at 22:13

$\delta > 0$ is not given but determined by $\delta_1$ and $\delta_2$ --- this is the correct logic. I will modify your wording of your proof:
Given $\varepsilon > 0$, there exists $\delta_1 > 0$ such that $|x - x_0| < \delta_1$ implies $|f(x) - L| < \frac{1}{2}\varepsilon$. Also, there exists $\delta_2 > 0$ such that $|x - x_0| < \delta_2$ implies $|g(x) - M| < \frac{1}{4}\varepsilon$.
Now take $\delta = \min(\delta_1, \delta_2) > 0$, then for all $x$ such that $|x - x_0| < \delta$, we have $$|f(x) + 2g(x) - (L + 2M)| \leq |f(x) - L| + 2|g(x) - M| < \frac{1}{2}\varepsilon + \frac{1}{2}\varepsilon = \varepsilon.$$ and the proof is complete.