## swin2013 3 years ago Limit problem?

1. swin2013

$\lim_{x \rightarrow -3^{+}} 1/x+3$

2. across

You have that$\lim_{x\to-3^+}\frac1{x+3}.$Immediate substitution yields a division by zero. Therefore, you must have a vertical asymptote there. Approaching it from the right side, however, will yield $$\infty$$ as the limit. On the other hand, approaching it from the left side will yield $$-\infty$$ as the limit.

3. ksaimouli

i am soory i read it wrong

4. swin2013

since it is on the right side, do i write that the limit is approaching positive infinity?

5. ksaimouli

actually the limit DNE because the absolute value get very large

6. across

Because it's a sided limit, you have to specify $$\text{where}$$ it goes. Simply writing DNE won't suffice.

7. viniterranova

I got it 1/6. Just plug it into the equation

8. across

@viniterranova, how did you get $$1/6$$?

9. swin2013

I'm pretty sure i should get either - infinity or positive. I'm kind of unclear on what to write. I understand that when i divide the constant (1) by x-2 (which is a small positive number), does that mean it's approaching positive infinity since 1 / x-2 is getting positively bigger?

10. viniterranova
11. swin2013

@viniterranova it's actually negative 3. it appears that the equation plugged in 3. if we plug in the value of -3, we get 0 on the denominator which is not valid since the denominator can't be zero

12. across

@swin2013, when you're asked to compute$\lim_{x\to-3^+}\frac1{x+3},$if you substitute the limit, $$-3$$, you'll have that$\frac1{x+3}=\frac1{(-3)+3}=\frac10.$Therefore, you know that the limit DNE. However, if you approach $$-3$$ from the right side, you'll have that$\frac1{x+3}$gets infinitely larger (since you're dividing by a really small number) toward $$\infty$$. Oh, and @viniterranova, the limit is $$-3$$, not $$3$$. Read the question again.

13. swin2013

so am i correct in the sense that x - infinity?

14. viniterranova

See teh wolfram site.

15. viniterranova

The fact of you have the minus sign after 3 doesn´t mean that the 3 is negative. Just mean that you are approaching from left.

16. viniterranova

Just this.

17. across

No, you silly, $$-3$$ is different from $$3^-$$.

18. swin2013

no. if it's approaching 3 to the left, there will be a negative as an exponent

19. across

@swin2013, for example, let $$k=-2.99999$$ be that close to $$-3$$. Then$\frac1{k+3}=\frac1{-2.99999+3}=\frac1{0.00001}=100,000.$So you indeed approach $$\infty$$.

20. across

(Notice that we're close to it from the right side.)

21. swin2013

thank you @across, i just need clarity on what to write for my answer haha. I already understand the concept :)

22. viniterranova

Across you didn´t understand me. I just said minus sign after 3 not before.

23. swin2013

yea but the equation never stated a negative sign after the 3.

24. across

That doesn't change the fact that you screwed up, @viniterranova. ;) @swin2013, you can write: "The limit does not exist as it approaches positive infinity."

25. across

You can also just write$\lim_{x\to-3^+}\frac1{x+3}=\infty.$It depends on your teacher.

26. swin2013

x - infinity no finite lim :)

27. ksaimouli

the correct way to say according to textbook is it DNE

28. ksaimouli

xD

29. across

lol @ksaimouli, you are right, but this is a SIDED LIMIT. In other words, YOU MUST SPECIFY WHERE IT IS GOING. <:)

30. swin2013

lol i never said INFINITE i said FINITE

31. swin2013

which is the same as DNE as in there is not possible limit

32. ksaimouli

ok

33. swin2013

or DEFINITE limit (the derivative of FINITE)