



Junior Member


Jul 13, 2009, 05:58 AM


Topic: infinite limits
Can anyone help me with this problem. :)
The given is: g(x) = x/(x5) and I need to find the vertical asymptote of the graph of g and the sketch of the graph of g for values close to the asymptote.
Here is how I solved it.
g(x) = x/(x5)
x>5
= (lim x) / (lim x  lim 5)
x>5 x>5
= 5/(55)
= 5/0
= positive infinity
I don't know if my answer is correct but I tried my best to answer it. I don't have a graph yet. :(
I hope you can help me with this. Thanks in advance! :)



Uber Member


Jul 13, 2009, 06:14 AM


You have to find at which value of x does g(x) equals to positive infinity. That value is the line where you have the vertical asymptote.
The graphing should be quite easy.



Junior Member


Jul 13, 2009, 06:25 AM


Wait. I want to know first if my answer 5/0 = positive infinity correct? :)



Uber Member


Jul 13, 2009, 06:28 AM


Yup, so what you think the value of x is? (you already got that! ;))



Junior Member


Jul 13, 2009, 06:32 AM


Is x=5?? :D



Uber Member


Jul 13, 2009, 06:34 AM


YES!! Congrats! :)



Junior Member


Jul 13, 2009, 06:43 AM


Thank you for your help! Clap! Clap! Clap! :D
wait. To get the values close to the asymptote.. Am I just going to substitute the value of x..
example: g(x) = (5.01)/(5.015)
is this correct? :) or not? :(



Uber Member


Jul 13, 2009, 06:52 AM


You can do that. However, if you're plotting, that might be somewhat difficult.
If you did the , that gives 501. Some larger numbers may perhaps be more practical. Still, I don't know how you do in your country to do such numbers, but you'll see that the graph decreases very fast, then it becomes more horizontal around x=8



Junior Member


Jul 13, 2009, 06:59 AM


Oh! I see. Thanks a lot! :D



Uber Member


Jul 13, 2009, 07:03 AM


And you can try plotting the part to the right of the asymptote. You'll get something that you have perhaps never seen when you go to the negative values of x.



Junior Member


Jul 13, 2009, 07:14 AM


Ok! Ok! I'll do that. :)
Hey! :D What do you think in this problem? Is it positive or negative infinity?
lim (x/x+2) + (1/x^24)
x>2^+
when I tried to answer it. I end with + infinity.



Junior Member


Jul 13, 2009, 07:21 AM


Ok! Ok! I'll do that. :)
Hey! :D What do you think in this problem? Is it positive or negative infinity?
lim (x/x+2) + (1/x^24)
x>2^+
when I tried to answer it. I end with + infinity.



Junior Member


Jul 13, 2009, 07:26 AM


My answer is wr0ng! :DD
It should be  infinity..



Uber Member


Jul 13, 2009, 07:28 AM


Wait a sec. Does that mean you and akotoh are the same person? I've seen the exact question on another recent thread :confused:
Anyway, the question just got answered.



Junior Member


Jul 13, 2009, 07:40 AM


Nope! I have read his/her post and I'm just confused what really the answer is. Because the other one says that the answer is  and the other says that its +.
That's it! :)



Uber Member


Jul 13, 2009, 07:46 AM


Just had the graph plotted. It goes to ve infinity, with asymptote at x = 2.



Junior Member


Jul 13, 2009, 07:58 AM


Hey! I've tried to substitute the value of x to g(x)= x/(x5)..
if..
g(6) = 6
g(7) = 3.5
g(8) = 2.67
g(9) = 2.25
g(10) = 2
am I correct? :D



Uber Member


Jul 13, 2009, 08:00 AM


Yup! :)



Junior Member


Jul 13, 2009, 08:05 AM


Okey! I'm totally done! :D
Thank you very much for your help!
I need to sleep now. Its already 11pm here in our country.
Hope you can help me again next time! Thanks! More powers! ;)



Uber Member


Jul 13, 2009, 08:07 AM


Ok, Good night! :)


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