## hawkfalcon 3 years ago Another lim problem, but different:(

1. hawkfalcon

$f(x) = \left[\begin{matrix}x^2-1 & x<1 \\ 4-x & x>=1\end{matrix}\right]$

2. hawkfalcon

I can do the lim, but how do i do it on that:O it lin x->1

3. hawkfalcon

$\lim_{x \rightarrow 1}$

4. Denebel

|dw:1346015959313:dw| Your graph will look something like this

5. Denebel

What is the limit for the function x^2 - 1 as it approaches 1?

6. hawkfalcon

Erm. -inf. right?

7. Denebel

Hmm, not quite. You want to approach x = 1.

8. Denebel

Follow the graph of x^2 -1 as it approaches x =1. What is the y-value?

9. hawkfalcon

0

10. hawkfalcon

I think:3

11. hawkfalcon

Sorry *.* i'm not very good a limits:(

12. Denebel

Yes :) then follow the graph of 4-x as it approaches x=1. You start from the right hand side and move back till you reach x=1.

13. Denebel

What would you say the limit was for that equation?

14. hawkfalcon

3

15. Denebel

Ok, are the two limits for both equations the same?

16. hawkfalcon

Nope. Oh:3. Dur dur. You compare the 2 equations:3

17. Denebel

18. hawkfalcon

DNE

19. hawkfalcon

@Denebel one last Q. It also says to find x->+ and -. In this cast, thats DNE too. right?

20. Denebel

Nope, as x->+ means when x approaches a number (in this case, 1) from the right hand side. x->- means when x approaches (1) from the left hand side.

21. hawkfalcon

But those are not the same for either left or right.

22. Denebel

You only need to make sure they are the same when they don't give you a +/- Since the next part of the problem asks you to state the limit when it comes from +, then when it comes from - you say what you told me before. So when x-> 1+ is when it approaches x=1 from the right side. You only look at the function 4-x in this case. x-> 1- is when it approaches x=1 from the left side, so look at the function x^2 - 1

23. hawkfalcon

Oh so you only look at each problem separately then :O

24. hawkfalcon

so lim x->1- = 0 and + = 3 and x-> = DNE

25. Denebel

Yup.

26. hawkfalcon

<3 Thank you so much!