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Evaluate the limit: can someone explain me this?
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 8 months ago
 8 months ago
Evaluate the limit: can someone explain me this? (attached below)
 8 months ago
 8 months ago

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mathcalculusBest ResponseYou've already chosen the best response.0
when we evaluate the limit, do we not substitute?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
this is where i'm confused.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
in this case, i got the answer right. i kind of took a smart guess
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
at x=11/8 the bottom goes to 0, right? But coming from the right, the bottom is neg, as is the top so it goes to inf. Coming from the left, the bottom is pos and the top is neg so it goes to minf. Make sense?
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
When the limit is NOT giving you an indeterminate form, yes you're allowed to substitute the value directly in :)
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@pgpilot326 a little
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
indeterminate form meaning.... 0/0?
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
You're really just testing points on each side of your undefined point. You need to see what the behavior is on each side, so pick a point on each side and see what it's doing.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i got the numerator..319/8 and the denominator 77/8
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
And indeterminate form can mean more than just 0/0. For the purposes of calc 1, yeah, 0/0
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i simply subtracted and simplified.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
but i saw that they asked infinity or non infinity etc etc
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Yeah, because it will go to infinity or negative infinity. You're approaching an asymptote, not an actual point. The graph will just shoot up or down. But the way to see whether it shoots up or down is to test points on each side of the asymptote.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so what i did was wrong?
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Kinda, yeah. I would just pick a point really close to the asymptote on each side. If it's really close, your number will either be really high in the positive direction or really high in the negative direction.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
so.... how should i do it?
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Like I picked the number 5/4 for the left side and got 145/4. Picking 3/2 for the right side I got 87/2. You can kinda see how one gets high fast and one goes low fast.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
im more confused now. :/
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
I would just personally pick numbers on each side of the asymptote that are very close to it. If they're close enough, the answer should tell you pretty clearly if you're going way negative or way positive. Maybe the others can explain better x_x
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yeah what if i wanted to use 11/8
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
it's kidn of the same thing... right?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i see that one goes really high up and the other goes really down.
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Then you will get an undefined answer. An undefined answer does you no good.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
no i didn't get an undefined answer...
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Oh, I apologize, I did not look correctly. I'm brain cramping then, ignore what I've said x_x
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
look, when x is near 11/8 but more, 11  8x is gonna be negative. neg over neg is pos. when x is near 11/8 but less, 11  8x is gonna be pos. neg over pos is neg. In both cases, 11  8x is headed towards 0 as x goes to 11/8.
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Wait, what? You DIDNT get an undefined? 11 8(11/8)?
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
I guess I was right and confused myself _ Well, my advice would to just do what I tried to explain. If you can test points on both sides of the asymptote and see theyre really positive or really negative, that you should tell you your answer. As to how you didn't get an undefined when you plugged in 11/8, I'm not sure?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@pgpilot326 show it pls
 8 months ago

PsymonBest ResponseYou've already chosen the best response.0
Yeah. nvm, I'm clearly being confusing. I'll leave, sorry.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
#1) can i simply substitute 11/8 into the equation? yes or no?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay no, this method confused me from the beginning. :/
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@zepdrix wait what?
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
lemme erase all that nonsense and see if i can explain this :)
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large \lim_{x\to11/8^+} \frac{29x}{118x} \qquad\to\qquad \frac{39.875}{0}\] Plugging the fraction directly in shows us that we're approaching this form. Since we're approaching a zero in the denominator, it means there's an asymptote at x=11/8. We need to figure something out though. Are we approaching positive or negative infinity? Well our numerator is negative. How about our denominator? When x is a tiny bit bigger than 11/8, is the denominator negative or positive?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@zepdrix how'd you get that number?
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
I plugged 11/8 directly in.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
oh ok i see where you got the number from
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
negative? Is that in response to the question about the denominator?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i think, not sure. i think it's negative.
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Yes, very good! Because \(\large 8x\) is slightly larger than \(\large 11\) when we plug in a value larger than 11/8. So \(\large 118x\) is giving us a negative value when we approach from the larger side.
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
think of the function 1/x. when x goes to zero, the function goes to inf from the right because x is positive and it goes to negative infinite from the left because x is negative. the limit doesn't exist because it goes to 2 different places when coming from the left and right. You're problem is the same except it goes to 0 on the bottom when x =11/8. So the limit won't exist at x=11/8 if the limit goes to 2 different places coming from the left and right. 118x=0 when x=11/8. But if x>11/8, 118x is negative and if x<11/8, 118x is positive. This is how you determine where it's heading as you approach the zero from either side.
 8 months ago

zepdrixBest ResponseYou've already chosen the best response.0
^ yah make sure you're comfortable with \(\large \lim_{x\to0}\dfrac{1}{x}\) as pilot explained.
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
But don't forget what the top is doing too.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay, so what are the exact rules to every limit problem. keeping in mind y/x if denominator is 0 then it's negative?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i know this: ex: 1/x all real numbers except x=0
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
with this problem. im just so so confused on to how to approach it.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
ex: step by step. and the understanding part is the hardest part.
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
\[\lim_{x \rightarrow a}f \left( x \right)=c \iff \lim_{x \rightarrow a ^{+}}f \left( x \right)=\lim_{x \rightarrow a ^{}}f \left( x \right)=c\] This says the limit exists if and only if the limit from the left and right are the same. You can have different left and right limits, in which case the overall limit will not exist. For example, let \[f \left( x \right)=\frac{ 1 }{ \left( x3 \right) }\] in order to determine \[\lim_{x \rightarrow 3}f \left( x \right)\] we need to look at what happens on the left and right as x approaches 3. \[\lim_{x \rightarrow 3^{+}}\frac{ 1 }{ \left( x3 \right) }=+\infty\]because \[\left( x3 \right)>0, \forall x>3\]. Likewise, \[\lim_{x \rightarrow 3^{}}\frac{ 1 }{ \left( x3 \right) }=\infty\]because \[\left( x3 \right)<0, \forall x<3\]. The bottom is going to 0 as x gets closer and closer to 3. But coming from above 3 (like 3.1, 3.01, 3.001), the bottom is still positive as it goes to zero. yetcoming from below 3 (like 2.9, 2.99, 2.999) the bottom is still negative as it goes to zero. I hope that helps cause it was alot to type.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
thanks i understand it much better.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i see that the left side is coming from the negative side. and right from the positive.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
now let's say for this problem:
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
For all of these evaluate the top and bottom separately and them bring them together. the bottom is always positive, no matter what side you come from. That's because you're squaring the bottom. The top is always going to be negative. Thus all of them will be negative infinity. Do you see?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i substituted all 7 in.... the problem is the signs right and left.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i got them wrong.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i typed in n, p, dne
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.1
You can break up the limit into the limit of the top times the limit of the bottom. If the limit exists in each case, then all of the limits (left, right and overall) will be the same. \[\lim_{x \rightarrow 7}\left( \frac{ 2x+1 }{ \left( x7 \right)^{2} } \right)=\lim_{x \rightarrow 7}\left( 2x+1 \right) \times \lim_{x \rightarrow 7}\left( \frac{ 1 }{ \left( x7 \right)^{2} } \right)\] \[\lim_{x \rightarrow 7}2x+1 = 13 \iff \lim_{x \rightarrow 7^{}}2x+1 =\lim_{x \rightarrow 7^{+}}2x+1 = 13\] Also \[\lim_{x \rightarrow 7}\left( \frac{ 1 }{ \left( x7 \right)^{2} } \right)=+\infty \iff \lim_{x \rightarrow 7^{}}\left( \frac{ 1 }{ \left( x7 \right)^{2} } \right)=\lim_{x \rightarrow 7^{+}}\left( \frac{ 1 }{ \left( x7 \right)^{2} } \right)=+\infty\] So \[\lim_{x \rightarrow 7}\left( \frac{ 2x+1 }{ \left( x7 \right)^{2} } \right)=\infty \iff \lim_{x \rightarrow 7^{}}\left( \frac{ 2x+1 }{ \left( x7 \right)^{2} } \right)=\lim_{x \rightarrow 7^{+}}\left( \frac{ 2x+1 }{ \left( x7 \right)^{2} } \right)=\infty\]
 8 months ago
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