## anonymous 2 years ago 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.

1. anonymous

2. anonymous

3. anonymous

is the answer $$e^{ab}$$?

4. anonymous

Yes

5. anonymous

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

6. anonymous

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???

7. anonymous

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

8. anonymous

the solution you have written is only algebra

9. anonymous

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

10. anonymous

that part is clear right?

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

Oh yes, sorry - z=a/x

15. anonymous

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

16. anonymous

OK

17. anonymous

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

18. anonymous

yep

19. anonymous

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

20. anonymous

ok - I see that now.

21. anonymous

done right?

22. anonymous

or are there still other algebra steps?

23. anonymous

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!!

24. anonymous

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

25. anonymous

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