Here's the question you clicked on:
swin2013
Limit problem?
\[\lim_{x \rightarrow -3^{+}} 1/x+3\]
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.
i am soory i read it wrong
since it is on the right side, do i write that the limit is approaching positive infinity?
actually the limit DNE because the absolute value get very large
Because it's a sided limit, you have to specify \(\text{where}\) it goes. Simply writing DNE won't suffice.
I got it 1/6. Just plug it into the equation
@viniterranova, how did you get \(1/6\)?
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?
http://www.wolframalpha.com/input/?i=limit+x-%3E+3%2B+%281%2F%28x%2B3%29
@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
@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.
so am i correct in the sense that x - infinity?
See teh wolfram site.
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.
No, you silly, \(-3\) is different from \(3^-\).
no. if it's approaching 3 to the left, there will be a negative as an exponent
@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 \(\infty\).
(Notice that we're close to it from the right side.)
thank you @across, i just need clarity on what to write for my answer haha. I already understand the concept :)
Across you didn´t understand me. I just said minus sign after 3 not before.
yea but the equation never stated a negative sign after the 3.
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."
You can also just write\[\lim_{x\to-3^+}\frac1{x+3}=\infty.\]It depends on your teacher.
x - infinity no finite lim :)
the correct way to say according to textbook is it DNE
lol @ksaimouli, you are right, but this is a SIDED LIMIT. In other words, YOU MUST SPECIFY WHERE IT IS GOING. <:)
lol i never said INFINITE i said FINITE
which is the same as DNE as in there is not possible limit
or DEFINITE limit (the derivative of FINITE)