## anonymous 5 years ago lim (1+1/x)^x x-> The answer is e but I do not know how to get there

1. amistre64

well, that is the definition for 'e' :)

2. anonymous

I do not understand what you mean

3. anonymous

sorry I did not mean to repeat

4. amistre64

i mean that 'e' is defined as the limit as x -> inf; (1+1/x)^x

5. amistre64

try using the definition of a derivative..

6. amistre64

f(x+h) - f(x) lim h->inf; ----------- h

7. anonymous

ok, I will try it

8. anonymous

one definition is it is the solution to $\int^x_1 \frac{1}{t}\,dt = 1$

9. anonymous

@amistre64: Its not right, the definition is $h \rightarrow 0$ As to make the secant a tangent!

10. amistre64

lol...same diff :)

11. anonymous

Ok, thanks amistre64, but I am not grasping the point

12. anonymous

What do you mean ami

13. amistre64

i mean.... i might be wrong ;)

14. amistre64

what math is this for hix?

15. anonymous

calculus I

16. anonymous

another definition of e is $e= \lim_{x\to\infty} (1+\frac{1}{x})^x$ in which case the answer is immediate.

17. amistre64

well, calc1 at this semester should be talking about limits right?

18. anonymous

i have a final tomorrow

19. anonymous

or you can take the log, compute the limit, get 1, and then 'exponentiate' to get e

20. amistre64

hmm.... my finals were 2 weeks ago; just starting the summer semester here

21. anonymous

I have to go to class, but please post anything that can help, and I will check it later. Thank you all very much. see ya

22. anonymous

$lim_{x\to\infty} (1+\frac{1}{x})^x$ $ln(1+\frac{1}{x})^x = x ln(1+\frac{1}{x})$ which is of the form $\infty \times 0$ rewrite as $\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}$

23. anonymous

Then use l'hopital. But this is a lot of unnecessary work since you have the definition of e to begin with.

24. anonymous

Let $y = \lim_{x\to\infty}(1+1/x)^x$ we have to put it in indeterminate form in order to employ l'hopital's rule so we take natural log of both sides $y = \lim_{x\to\infty}(1+1/x)^x \rightarrow \ln{y} = \lim_{x\to\infty}\ln((1+1/x)^x)$ Using log rule of exponents we have $\lim_{x\to\infty}x*\ln((1+1/x))\rightarrow \infty-0$ Rewriting the whole expression $\lim_{x\to\infty}\ln((1+1/x))/(1/x)\rightarrow 0/0$ which is indeterminate, using L'Hopital's rule we have $\lim_{x\to\infty}(-x/(x^2(x+1))/(-1/x^2)\rightarrow \lim_{x\to\infty} x/(x+1) \rightarrow \infty/\infty$ which permits us to use L'Hopital's rule again so we have $\ln{y} = \lim_{x\to\infty}(1/1) = 1$ Raising both sides to e yields $y = e^1=e$ Done.