Evaluating Limit: (1 + \frac{1}{x})^x = e

[whatlifeforme]

Homework Statement

Evaluate.

Homework Equations

$lim_{x->infinity} (1 + \frac{1}{x})^x = e$

The Attempt at a Solution

lne = x(1+$\frac{1}{x})$

$\frac{1+\frac{1}{x}}{1/x}$ = 1/0 = ∞ (is this correct?

 

[Mark44]

Homework Statement

Evaluate.

Homework Equations

$lim_{x->infinity} (1 + \frac{1}{x})^x = e$

The Attempt at a Solution

lne = x(1+$\frac{1}{x})$

$\frac{1+\frac{1}{x}}{1/x}$ = 1/0 = ∞ (is this correct?

No, and you're going about it the wrong way, it seems to me. You're supposed to evaluate the given limit, and show that it is e.

One way to go is to let y = $\lim_{x->\infty} (1 + \frac{1}{x})^x$, then take ln of both sides. If you work with it a bit, you can get something that you can use L'Hopital's Rule on.

BTW, in LaTeX, ∞ is \infty, and limit is \lim.

 

[jbunniii]

Homework Statement

Evaluate.

Evaluate what?

Homework Equations

$lim_{x->infinity} (1 + \frac{1}{x})^x = e$

The Attempt at a Solution

lne = x(1+$\frac{1}{x})$

$\frac{1+\frac{1}{x}}{1/x}$ = 1/0 = ∞ (is this correct?

No, these are both nonsense. Surely you don't mean to claim that these equations hold for all $x$?

 

[whatlifeforme]

No, and you're going about it the wrong way, it seems to me. You're supposed to evaluate the given limit, and show that it is e.

One way to go is to let y = $\lim_{x->\infty} (1 + \frac{1}{x})^x$, then take ln of both sides. If you work with it a bit, you can get something that you can use L'Hopital's Rule on.

BTW, in LaTeX, ∞ is \infty, and limit is \lim.

if i take ln i get x(1+1/x) -->> make form for L'Hopital's rule: $\frac{1+(1/x)}{1/x}$

which is 1/0 --> applying l'hopital's again just makes it more complicated but no closer to the answer needed.

 

[whatlifeforme]

if i apply L'H again i get : (-1/x^2)/(-1/x^2) which would = 1 but the limit of that would be zero correct?

if the limit = 1 then the that makes the answer e.

 

[Mark44]

if i take ln i get x(1+1/x)

You're leaving something out.
If y = (1 + 1/x)^x,
then ln(y) = ln[(1 + 1/x)^x]
Now work with the right side. ln should not go away.

-->> make form for L'Hopital's rule: $\frac{1+(1/x)}{1/x}$

which is 1/0 --> applying l'hopital's again just makes it more complicated but no closer to the answer needed.

 

[HallsofIvy]

if i take ln i get x(1+1/x)

NO, you don't. You get x ln(1+ 1/x)

-->> make form for L'Hopital's rule: $\frac{1+(1/x)}{1/x}$

which is 1/0 --> applying l'hopital's again just makes it more complicated but no closer to the answer needed.

 

[whatlifeforme]

NO, you don't. You get x ln(1+ 1/x)

sorry. well even if that is the case:

lim (X->∞) $\frac{ln(1+(1/x)}{1/x}$ = 0/0


then

L'Hopitals -->> $\displaystyle \frac{\frac{1}{1+(1/x)}}{\frac{-1}{x^2}}$ = 1/0

1/0 still keeps popping up; do i conclude that the limit of lny = ∞

1/0 is not an indeterminate form in which l'hopital's can be applied, correct?

furthermore, if we conclude that 1/0 = ∞

then we have:

lim (x->∞) lny = ∞
lim (x->∞) $e^{lny}$ = $e^∞$

but I'm looking for $e^1$ not $e^∞$

 

[whatlifeforme]

i found some help on another website:

lim_(x->oo)(1 + 1/x)^x = lim_(x->0)(1 + x)^(1/x)`

Let `y = (1 + x)^(1/x)`

how does lim_(x->oo)(1 + 1/x)^x = lim_(x->0)(1 + x)^(1/x)`

 

[phosgene]

sorry. well even if that is the case:

lim (X->∞) $\frac{ln(1+(1/x)}{1/x}$ = 0/0then

L'Hopitals -->> $\displaystyle \frac{\frac{1}{1+(1/x)}}{\frac{-1}{x^2}}$ = 1/0

What does the chain rule say when differentiating a function f(g(x))? Think about that and then try to re-do the step where you apply L'Hopital's rule.