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BrighterDays
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I have the answer....I just don't understand the steps. Find the limit as x approaches infinity of 1+(a/x)^(bx)  I am putting the original question and the answer from my professor with his explanation in the comments. I just don't understand the steps at all that he has written out.
 4 months ago
 4 months ago
BrighterDays Group Title
I have the answer....I just don't understand the steps. Find the limit as x approaches infinity of 1+(a/x)^(bx)  I am putting the original question and the answer from my professor with his explanation in the comments. I just don't understand the steps at all that he has written out.
 4 months ago
 4 months ago

This Question is Closed

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
is the answer \(e^{ab}\)?
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
you can guess it are you supposed to use l'hopital?
 4 months ago

BrighterDays Group TitleBest ResponseYou've already chosen the best response.0
In the comments I put a .png of how he worked it out....I don't understand his substitution with z and how he goes from there....he did say you could use l'hopital but it would be more work???
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
ooh i see the solution now, sorry you are supposed to use algebra, and then the fact that \[e=\lim_{x\to 0}\left(1+x\right)^{\frac{1}{x}}\]
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
the solution you have written is only algebra
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
they say put \(z=\frac{a}{x}\) so the piece inside is now \(1+z\)
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
that part is clear right?
 4 months ago

BrighterDays Group TitleBest ResponseYou've already chosen the best response.0
ugh. OK. so what I don't get is the algebra of: this: if z=a/b how does bx=abz^1?
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
the exponential notation may have confused you lets write it without the exponential notation
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
btw it is not \(z=\frac{a}{b}\) but rather \(z=\frac{a}{x}\) right?
 4 months ago

BrighterDays Group TitleBest ResponseYou've already chosen the best response.0
Oh yes, sorry  z=a/x
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
ok step by step \[\large z=\frac{a}{x}\] solve for \(x\) you get \[\large x=\frac{a}{z}\] ok so far?
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
multiply both sides by \(b\) you get \[\large bx=\frac{ab}{z}\] right?
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
well that is it then \[\left(1+\frac{a}{x}\right)^{bx}\] becomes \[\large (1+z)^{\frac{ab}{z}}\]
 4 months ago

BrighterDays Group TitleBest ResponseYou've already chosen the best response.0
ok  I see that now.
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
done right?
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
or are there still other algebra steps?
 4 months ago

BrighterDays Group TitleBest ResponseYou've already chosen the best response.0
That's it! Thank you. So the fact that e= limx→0 of (1+x)^1/z is just something we need to know, correct? I keep looking for it in my book but can't find it. So, I'll just write it and make a note of it. Thanks again!!
 4 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
well i guess so usually it is written as \[e=\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x\] but if you change \(x\) to \(\frac{1}{x}\) then you get \[e=\lim_{x\to 0}(1+x)^{\frac{1}{x}}\]
 4 months ago

BrighterDays Group TitleBest ResponseYou've already chosen the best response.0
Great! Thank you so much for all your help!!
 4 months ago
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