## swin2013 Group Title Limit problem? one year ago one year ago

1. swin2013 Group Title

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

2. across Group Title

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 Group Title

i am soory i read it wrong

4. swin2013 Group Title

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

5. ksaimouli Group Title

actually the limit DNE because the absolute value get very large

6. across Group Title

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

7. viniterranova Group Title

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

8. across Group Title

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

9. swin2013 Group Title

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 Group Title
11. swin2013 Group Title

@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 Group Title

@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 Group Title

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

14. viniterranova Group Title

See teh wolfram site.

15. viniterranova Group Title

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 Group Title

Just this.

17. across Group Title

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

18. swin2013 Group Title

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

19. across Group Title

@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 Group Title

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

21. swin2013 Group Title

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

22. viniterranova Group Title

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

23. swin2013 Group Title

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

24. across Group Title

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 Group Title

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

26. swin2013 Group Title

x - infinity no finite lim :)

27. ksaimouli Group Title

the correct way to say according to textbook is it DNE

28. ksaimouli Group Title

xD

29. across Group Title

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

30. swin2013 Group Title

lol i never said INFINITE i said FINITE

31. swin2013 Group Title

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

32. ksaimouli Group Title

ok

33. swin2013 Group Title

or DEFINITE limit (the derivative of FINITE)