Question

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

Answer 1

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

Answer 2

I do not understand what you mean

Answer 3

sorry I did not mean to repeat

Answer 4

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

Answer 5

try using the definition of a derivative..

Answer 6

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

Answer 7

ok, I will try it

Answer 8

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

Answer 9

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

Answer 10

lol...same diff :)

Answer 11

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

Answer 12

What do you mean ami

Answer 13

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

Answer 14

what math is this for hix?

Answer 15

calculus I

Answer 16

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

Answer 17

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

Answer 18

i have a final tomorrow

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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.