Question

Prove that if lim(x→c) f(x) = L then lim(x→c) 7 f(x) = 7L.

Answer 1

Hey,
Welcome to Open Study! :)

What do they mean by prove?
Using Epsilon-Delta definition of a limit?

Or just basic Limit Properties?

Answer 2

Using the Epsilon-Delta definition.

Answer 3

Hmm ok lemme see if my line of thinking makes sense here...

Answer 4

\(\large\rm \lim\limits_{x\to c}~f(x)=L\) means that

for all \(\rm \epsilon>0\) there exists a \(\rm \delta>0\) such that
\(\rm 0<|x-c|<\delta\) when x is within delta of c,
\(\rm |f(x)-L|<\epsilon\) then f(x) is within epsilon of L.

Answer 5

Hence when \(\large\rm |f(x)-L|<\epsilon\)
it should follow that \(\large\rm 7|f(x)-L|<7\epsilon\),
and \(\large\rm |7f(x)-7L|<7\epsilon\).

Answer 6

Let \(\large\rm \delta=7\epsilon\) so that when
\[\large\rm 0<|x-c|<\delta\]we have\[\large\rm |7f(x)-7L|<\epsilon^*\](Where epsilon*=7epsilon).
This definition should hold `for all` epsilon, ya?

And then relate it back to the definition of limit I guess?
Therefore, by definition of limit,\[\large\rm \lim_{x\to c}7f(x)=7L\]

Answer 7

Am I doing that right? >.<
Hmm thinkinggg

Answer 8

Looks good so far.

Answer 9

Mmm ya I think that makes sense.
We have some new epsilon, a larger epsilon, \(\rm 7\epsilon\),
and we're simply showing that we can still find a delta such that the definition holds true.

Ya I would go with something like that. :)
I know my notes are a little sloppy, hopefully you can make sense of that though.

Answer 10

Thanks, I appreciate it.