1
$\begingroup$

Question:

Let $a \in R$ . If $\lim_{x\to a} f(x) = 2$ and $\lim_{x\to a} g(x) = 9$, show a $\delta$-$\epsilon$ proof that $\lim_{x\to a} (\pi f(x)-10g(x)) = 2\pi -90$.

My attempt at it:

Let $\epsilon > 0, \exists \delta_1,\delta_2,\delta_3>0$ such that $$0<|x-a|<\delta_1 \implies |f(x)-2|<\epsilon$$ $$0<|x-a|<\delta_2 \implies |g(x)-9|<\epsilon$$ $$0<|x-a|<\delta_3 \implies |\pi f(x)-10g(x) - 2\pi -90|<\epsilon$$

Consider:

$$|\pi f(x)-10g(x) - 2\pi -90| = |\pi f(x)-10g(x) - 0 + 0 - 2\pi -90| = $$ $$|\pi (f(x)-2)- 0 + 0 - 10(g(x)-9)| $$ $$\le |\pi||f(x)-2| + |10||g(x)-9|$$

At this point I'm not sure how to continue. Do I just assume $|f(x)-2|<\epsilon$ and $|g(x)-9|<\epsilon$ and say that $$\le |\pi||f(x)-2| + |10||g(x)-9|$$ $$< |\pi|\epsilon + |10|\epsilon ?$$

Or do I need to figure out what $\delta_1$ and $\delta_2$ is to continue, but if that is the case how would I do that?

If anyone could help me with this that would be really appreciated.

$\endgroup$
2
$\begingroup$

You would be fine if you let $\delta$ be the smaller of $\delta_1$ and $\delta_2$, and note that for all $x$, $|x - a| < \delta$ implies $|x - a| < \delta_1$ and $|x - a| < \delta_2$, which implies $|f(x) - 2| < \epsilon$ and $|g(x) - 9| < \epsilon$; hence $$|\pi f(x) - 10 g(x) - (2\pi - 90)| = |\pi(f(x) - 2) - 10(g(x) - 9)| \le \pi|f(x) - 2| + 10|g(x) - 9| < (\pi + 10)\epsilon.$$ As $\epsilon$ was arbitrary, the result follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.