Limit problem?

- anonymous

Limit problem?

- schrodinger

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- anonymous

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

- 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.

- ksaimouli

i am soory i read it wrong

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## More answers

- anonymous

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

- ksaimouli

actually the limit DNE because the absolute value get very large

- across

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

- anonymous

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

- across

@viniterranova, how did you get \(1/6\)?

- anonymous

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?

- anonymous

http://www.wolframalpha.com/input/?i=limit+x-%3E+3%2B+%281%2F%28x%2B3%29

- anonymous

@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

- 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.

- anonymous

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

- anonymous

See teh wolfram site.

- anonymous

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.

- anonymous

Just this.

- across

No, you silly, \(-3\) is different from \(3^-\).

- anonymous

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

- 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 \(\infty\).

- across

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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."

- across

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

- anonymous

x - infinity no finite lim :)

- ksaimouli

the correct way to say according to textbook is it DNE

- ksaimouli

xD

- across

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

- anonymous

lol i never said INFINITE i said FINITE

- anonymous

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

- ksaimouli

ok

- anonymous

or DEFINITE limit (the derivative of FINITE)

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