## 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. 3 months ago 3 months ago

1. BrighterDays Group Title

2. BrighterDays Group Title

3. satellite73 Group Title

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

4. BrighterDays Group Title

Yes

5. satellite73 Group Title

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

6. BrighterDays Group Title

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. satellite73 Group Title

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. satellite73 Group Title

the solution you have written is only algebra

9. satellite73 Group Title

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

10. satellite73 Group Title

that part is clear right?

11. BrighterDays Group Title

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

12. satellite73 Group Title

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

13. satellite73 Group Title

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

14. BrighterDays Group Title

Oh yes, sorry - z=a/x

15. satellite73 Group Title

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

16. BrighterDays Group Title

OK

17. satellite73 Group Title

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

18. BrighterDays Group Title

yep

19. satellite73 Group Title

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

20. BrighterDays Group Title

ok - I see that now.

21. satellite73 Group Title

done right?

22. satellite73 Group Title

or are there still other algebra steps?

23. BrighterDays Group Title

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. satellite73 Group Title

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. BrighterDays Group Title

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