Use L’Hopital Rule to solve the limits: lim x-->0+ [1+(1/x)]^x?

Answer 1

It is e of course.

The one proof by jim though hardly legible was correct.

Answer 2

This limit is not equal to e. The limit is 1.

For Jim:
e = Lim x--> Infinity [1+x]^(1/x) = 2.718...
OR
e = Lim x->0 [1+x]^(1/x) = 2.718...

You can *always* check your limit answers by performing some basic arithmetic with numbers. For your example: Take very small values of x, say 0.0001 and then 0.000001 and plug these into your formula. You will notice that each time, the result gets closer and closer to 1.

Answer 3

This is a _very_ important limit and you should memorize the solution because you will almost certainly see a problem just like it on your calculus test! :)

Let
L=\lim_{x \to 0^+} (1+1/x)^x.

Take the natural logarithm of both sides:

ln( L)=\lim_{x \to 0^+} x ln (1+1/x)

Rewrite to obtain an inteterminate form of type
infinity/infinity:

ln( L)=\lim_{x \to 0^+} ln (1+1/x) / (1/x)

Apply L'Hospital's Rule:

ln( L)=\lim_{x \to 0^+} 1/(1+1/x) * -1/x^2 / (-1/x^2)

Take the limit:

ln (L) =1

Exponentiate to recover L:

L=e^(ln (L))=e^1=e.

The limit is e.

You should verify that
\lim_{x \to 0^+} (1+r/x)^x=e^(r/x) for all real r! :)

Answer 4

I would think 0 b/c if x is really high, 1/x is close to zero, so you virtually have 1^x, which will always be 1

Answer 5

let A=Lt.x->0[1+(1/x)]^x
log A=Lt.x->0 xlog[1+(1/x)]
=Lt.x->0 log[1+(1/x)]/1/x
=Lt.x->0{1/[1+(1/x)]}*(-1/x^2)/(-1/x^2)
=Lt x->0 x/x+1=0(answer)
therefor A=1 (answer)

Answer 6

xlog2x