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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, that is the definition for 'e' :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I do not understand what you mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry I did not mean to repeat

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i mean that 'e' is defined as the limit as x > inf; (1+1/x)^x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0try using the definition of a derivative..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f(x+h)  f(x) lim h>inf;  h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one definition is it is the solution to \[\int^x_1 \frac{1}{t}\,dt = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@amistre64: Its not right, the definition is \[h \rightarrow 0\] As to make the secant a tangent!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks amistre64, but I am not grasping the point

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i mean.... i might be wrong ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what math is this for hix?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0another definition of e is \[e= \lim_{x\to\infty} (1+\frac{1}{x})^x\] in which case the answer is immediate.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, calc1 at this semester should be talking about limits right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a final tomorrow

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or you can take the log, compute the limit, get 1, and then 'exponentiate' to get e

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hmm.... my finals were 2 weeks ago; just starting the summer semester here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[lim_{x\to\infty} (1+\frac{1}{x})^x\] \[ln(1+\frac{1}{x})^x = x ln(1+\frac{1}{x})\] which is of the form \[\infty \times 0\] rewrite as \[\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then use l'hopital. But this is a lot of unnecessary work since you have the definition of e to begin with.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let \[y = \lim_{x\to\infty}(1+1/x)^x\] 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 \[\lim_{x\to\infty}x*\ln((1+1/x))\rightarrow \infty0\] 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.
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