anonymous
  • anonymous
lim (1+1/x)^x x-> The answer is e but I do not know how to get there
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
well, that is the definition for 'e' :)
anonymous
  • anonymous
I do not understand what you mean
anonymous
  • anonymous
sorry I did not mean to repeat

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
i mean that 'e' is defined as the limit as x -> inf; (1+1/x)^x
amistre64
  • amistre64
try using the definition of a derivative..
amistre64
  • amistre64
f(x+h) - f(x) lim h->inf; ----------- h
anonymous
  • anonymous
ok, I will try it
anonymous
  • anonymous
one definition is it is the solution to \[\int^x_1 \frac{1}{t}\,dt = 1\]
anonymous
  • anonymous
@amistre64: Its not right, the definition is \[h \rightarrow 0\] As to make the secant a tangent!
amistre64
  • amistre64
lol...same diff :)
anonymous
  • anonymous
Ok, thanks amistre64, but I am not grasping the point
anonymous
  • anonymous
What do you mean ami
amistre64
  • amistre64
i mean.... i might be wrong ;)
amistre64
  • amistre64
what math is this for hix?
anonymous
  • anonymous
calculus I
anonymous
  • anonymous
another definition of e is \[e= \lim_{x\to\infty} (1+\frac{1}{x})^x\] in which case the answer is immediate.
amistre64
  • amistre64
well, calc1 at this semester should be talking about limits right?
anonymous
  • anonymous
i have a final tomorrow
anonymous
  • anonymous
or you can take the log, compute the limit, get 1, and then 'exponentiate' to get e
amistre64
  • amistre64
hmm.... my finals were 2 weeks ago; just starting the summer semester here
anonymous
  • 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
anonymous
  • 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 \[\frac{ln(1+\frac{1}{x})}{\frac{1}{x}}\]
anonymous
  • anonymous
Then use l'hopital. But this is a lot of unnecessary work since you have the definition of e to begin with.
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.