Re: Limits are limiting me

Hey M670.

Hint: You might want to consider a substitution where lim u -> g(x)- (i.e. from the left) and consider this in terms of f(u) instead of f(g(x)).

Re: Limits are limiting me

For #1, you can find the limit of g(x) and just look at the graph to see what f(g(x)) is there.

For #4, you need to think a little harder. Since x goes to 3 from above, g(x) goes to 0 from above. Can you see that in the graph? So now you look at the graph for f, and when its input goes to 0 from above, what does its output do?

- Hollywood

Re: Limits are limiting me

Oh I think I understand what I need to do, if I treat this like when doing compositions of function I figure out my g(x) and input it as my (x) in the f(g(x))

Ok but for #1 g(x) I am looking at the blue graph and my $\displaystyle x\rightarrow 0^- $ which means I am looking from right to left, this is where I get confused cuz it looks like its around 1 but then x @ 1 in the red graph gives us 0 on the y axis....

Re: Limits are limiting me

Quote:

Originally Posted by

**M670** Oh I think I understand what I need to do, if I treat this like when doing compositions of function I figure out my g(x) and input it as my (x) in the f(g(x))

Ok but for #1 g(x) I am looking at the blue graph and my $\displaystyle x\rightarrow 0^- $ which means I am looking from right to left, this is where I get confused cuz it looks like its around 1 but then x @ 1 in the red graph gives us 0 on the y axis....

Yes, in #1, as x goes to 0 g(x) goes to 1. And then, as g(x) goes to 1, f(g(x)) goes to 0. $\displaystyle \lim_{x\to 0^-} f(g(x))= 0$

Why are you confused about that?

Re: Limits are limiting me

Ok so then for #4 g(x) goes to 0 and the f(x) goes to 4 is my first thought but looking at $\displaystyle x \rightarrow 3^-$ it would seem to be approximately 2 on the y axis

Re: Limits are limiting me

Yes, that's right: x goes to 3 from above, so g(x) goes to 0 __from above__, since the graph of g(x) is above the x-axis just to the right of x=3. Since g(x) goes to 0 from above, f(g(x)) goes to 2.

- Hollywood