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.

Question

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.

anonymous · Answer

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anonymous · Answer

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anonymous · Answer

is the answer $e^{ab}$?

anonymous · Answer

Yes

anonymous · Answer

you can guess it
are you supposed to use l'hopital?

anonymous · Answer

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???

anonymous · Answer

ooh i see the solution now, sorry
you are supposed to use algebra, and then the fact that 
$$e=\lim_{x\to 0}\left(1+x\right)^{\frac{1}{x}}$$

anonymous · Answer

the solution you have written is only algebra

anonymous · Answer

they say put $z=\frac{a}{x}$ so the piece inside is now $1+z$

anonymous · Answer

that part is clear right?

anonymous · Answer

ugh. OK.  so what I don't get is the algebra of: this: if z=a/b how does bx=abz^-1?

anonymous · Answer

the exponential notation may have confused you 
lets write it without the exponential notation

anonymous · Answer

btw it is not $z=\frac{a}{b}$  but rather $z=\frac{a}{x}$ right?

anonymous · Answer

Oh yes, sorry - z=a/x

anonymous · Answer

ok step by step
$$\large z=\frac{a}{x}$$ solve for $x$ you get $$\large x=\frac{a}{z}$$ ok so far?

anonymous · Answer

OK

anonymous · Answer

multiply both sides by $b$ you get 
$$\large bx=\frac{ab}{z}$$ right?

anonymous · Answer

yep

anonymous · Answer

well that is it then 

$$\left(1+\frac{a}{x}\right)^{bx}$$ becomes 
$$\large (1+z)^{\frac{ab}{z}}$$

anonymous · Answer

ok - I see that now.

anonymous · Answer

done right?

anonymous · Answer

or are there still other algebra steps?

anonymous · Answer

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!!

anonymous · Answer

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 
$$e=\lim_{x\to 0}(1+x)^{\frac{1}{x}}$$

anonymous · Answer

Great! Thank you so much for all your help!!