1. ## Continuous limit

Can you tell me if I am on the right track on this problem

If f and g are continuous functions with f(-3) = 5 and the following limit, find g(-3).

$\displaystyle \lim_{x \to -3} [4f(x)-g(x)] = 13$

My work:

$\displaystyle \lim_{x \to -3}g(-3) = ?$

$\displaystyle 4\lim_{x \to -3}f(-3) = 5$

$\displaystyle g(-3) = 7$

I just winged this one since I don't know how to do it

Am I even close? lol

2. Originally Posted by silencecloak
Can you tell me if I am on the right track on this problem

If f and g are continuous functions with f(-3) = 5 and the following limit, find g(-3).

$\displaystyle \lim_{x \to -3} [4f(x)-g(x)] = 13$

My work:

$\displaystyle \lim_{x \to -3}g(-3) = ?$

$\displaystyle 4\lim_{x \to -3}f(-3) = 5$

$\displaystyle g(-3) = 7$

I just winged this one since I don't know how to do it

Am I even close? lol
i assume they are continuous on the real line

no, you're not close

$\displaystyle f(-3) = 5$, since $\displaystyle f(x)$ is continuous, $\displaystyle \lim_{x \to -3}f(x) = f(-3)$ (you should know this), thus $\displaystyle 4 \lim_{x \to -3}f(x) = 4 \cdot f(-3) = 4 \cdot 5 = 20$

3. $\displaystyle \lim_{x \to -3} [20-g(x)] = 13$

$\displaystyle 20-x = 13$

$\displaystyle x=7$

4. Originally Posted by silencecloak
$\displaystyle \lim_{x \to -3} [20-g(x)] = 13$

$\displaystyle 20-x = 13$

$\displaystyle x=7$
exactly how did you get x? how do you know $\displaystyle \lim_{x \to -3}g(x) = x$?

5. Originally Posted by Jhevon
exactly how did you get x?

$\displaystyle $$\displaystyle 4\lim_{x \to -3}f(-3) - \lim_{x \to -3}g(x) = 13 = \displaystyle 20 - g(x) = 13 6. Originally Posted by silencecloak \displaystyle$$\displaystyle 4\lim_{x \to -3}f(-3) - \lim_{x \to -3}g(x) = 13$

=

$\displaystyle 20 - g(x) = 13$
don't drop the limit from the g(x), you can't do that. you must solve for it

note that since g(x) is continuous on the reals, $\displaystyle \lim_{x \to -3}g(x) = g(-3)$, i told you the same thing for f.

7. Originally Posted by Jhevon
don't drop the limit from the g(x), you can't do that. you must solve for it

note that since g(x) is continuous on the reals, $\displaystyle \lim_{x \to -3}g(x) = g(-3)$, i told you the same thing for f.

So it would be

$\displaystyle 20 - \lim_{x \to -3}g(-3) = 13$

$\displaystyle \lim_{x \to -3}g(-3) = 7$

8. Originally Posted by silencecloak
So it would be

$\displaystyle$
$\displaystyle 20 - \lim_{x \to -3}g(-3) = 13$

$\displaystyle$
$\displaystyle \lim_{x \to -3}g(-3) = 7$

yes

do you see why? do you understand what's going on? write out your full solution

9. Originally Posted by Jhevon
yes

do you see why? do you understand what's going on? write out your full solution
$\displaystyle \lim_{x \to -3} [4f(x)-g(x)] = 13$

$\displaystyle \lim_{x \to -3}4f(x) - \lim_{x \to -3}g(x) = 13$

$\displaystyle 20 - \lim_{x \to -3}g(x) = 13$

$\displaystyle \lim_{x \to -3}g(x) = 7$

10. Originally Posted by silencecloak
$\displaystyle \lim_{x \to -3} [4f(x)-g(x)] = 13$

$\displaystyle \lim_{x \to -3}4f(x) - \lim_{x \to -3}g(x) = 13$

$\displaystyle 20 - \lim_{x \to -3}g(x) = 13$

$\displaystyle \lim_{x \to -3}g(x) = 7$
yes, and since g is continuous on the reals, $\displaystyle \lim_{x \to -3}g(x) = g(-3)$, thus $\displaystyle g(-3) = 7$

you must write that, otherwise, you have not answered the question. no one asked to for a limit, they asked you for g(-3)

11. Originally Posted by Jhevon
yes, and since g is continuous on the reals, $\displaystyle \lim_{x \to -3}g(x) = g(-3)$, thus $\displaystyle g(-3) = 7$

you must write that, otherwise, you have not answered the question. no one asked to for a limit, they asked you for g(-3)
Oh ok, that one hit home for me

Thanks for makin sure i understood it rather than just checking my answer

12. Originally Posted by silencecloak
Oh ok, that one hit home for me

Thanks for makin sure i understood it rather than just checking my answer
i'd also put in an extra line to show that $\displaystyle \lim_{x \to -3}4f(x) = 4f(-3)$. just so your professor knows you know what you're talking about