1. anonymous

2. anonymous

Need an explanation on how to do this.

3. anonymous

@phi

4. anonymous

@mathmath333

5. anonymous

@freckles

6. anonymous

@imqwerty

7. anonymous

@triciaal

8. Owlcoffee

We define continuity for a function "f" as the coicident lateral limit of the intermediate point. We can traduce like this: $\lim_{x \rightarrow a ^{\pm}} f(x)=f(a)$ What does this mean? In very simple words, the sufficient and necessary condition for a function to be continous at a point as long as the lateral limits of such point have a value coincident with the image of that point. Let's take a look at that and apply it to the problem in question, as you can see. Since the function is divided we will study what happens on that x=-4 point. So we will study the two following limits: $\lim_{x \rightarrow -4^{-}}\frac{ x+4 }{ x^2-16 }$ and $\lim_{x \rightarrow -4^{+}}\frac{ c }{ x+12 }$ These two limits arise from the very definition of limits, since when $$x<-4$$, the function f takes the structure of $$\frac{ x+4 }{ x^2-16 }$$ and when $$x \ge -4$$ the function f takes the structure $$\frac{ c }{ x+12 }$$. Now, solving those limits will give a value, and then you will want to use "c" as a variable to make them equal, therefore making the function continous.

9. anonymous

But when you plug in -4 they equal 0 so how can you equal both equations together?

10. anonymous

@Owlcoffee

11. Owlcoffee

Let's first work with: $\lim_{x \rightarrow (-4)^-} \frac{ x+4 }{ x^2-16 }=\frac{ 0 }{ 0 }$ This is an indetermination, which means that that you'll have to apply some factorization strategies in order to lift up the indetermination and solve the limit.

12. anonymous

So then it would be $\frac{ 1 }{ (x-4) }$ right?

13. anonymous

@Owlcoffee

14. Owlcoffee

Yes, correct, and then you plug "-4", what do you get?

15. anonymous

-1/8

16. anonymous

Then what do I do?

17. Owlcoffee

Good, now solve the second limit: $\lim_{x \rightarrow (-4)^+}\frac{ c }{ x+12 }$ In this case, just treat "c" as a constant and apply the limit.

18. anonymous

$\frac{ -1 }{ 8 }=\frac{ c }{ 8 }$$-8=8c$$-1=c$

19. anonymous

Is that right?

20. Owlcoffee

It is correct.

21. anonymous

Okay thanks!