- #1

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## Homework Statement

Evaluate.

## Homework Equations

[itex]lim_{x->infinity} (1 + \frac{1}{x})^x = e[/itex]

## The Attempt at a Solution

lne = x(1+[itex]\frac{1}{x})[/itex]

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

- Thread starter whatlifeforme
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- #1

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Evaluate.

[itex]lim_{x->infinity} (1 + \frac{1}{x})^x = e[/itex]

lne = x(1+[itex]\frac{1}{x})[/itex]

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

- #2

Mark44

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No, and you're going about it the wrong way, it seems to me. You're supposed to evaluate the given limit, and## Homework Statement

Evaluate.

## Homework Equations

[itex]lim_{x->infinity} (1 + \frac{1}{x})^x = e[/itex]

## The Attempt at a Solution

lne = x(1+[itex]\frac{1}{x})[/itex]

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

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.

- #3

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Evaluate what?## Homework Statement

Evaluate.

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

[itex]lim_{x->infinity} (1 + \frac{1}{x})^x = e[/itex]

## The Attempt at a Solution

lne = x(1+[itex]\frac{1}{x})[/itex]

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

- #4

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if i take ln i get x(1+1/x) -->> make form for L'Hopital's rule: [itex]\frac{1+(1/x)}{1/x}[/itex]No, and you're going about it the wrong way, it seems to me. You're supposed to evaluate the given limit, andshowthat 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.

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

- #5

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

- #6

Mark44

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You're leaving something out.if i take ln i get x(1+1/x)

If y = (1 + 1/x)

then ln(y) = ln[(1 + 1/x)

Now work with the right side. ln should not go away.

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

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

- #7

HallsofIvy

Science Advisor

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

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

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

- #8

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sorry. well even if that is the case:NO, you don't. You get x ln(1+ 1/x)

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

then

L'Hopitals -->> [itex]\displaystyle \frac{\frac{1}{1+(1/x)}}{\frac{-1}{x^2}}[/itex] = 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->∞) [itex]e^{lny}[/itex] = [itex]e^∞[/itex]

but i'm looking for [itex]e^1[/itex] not [itex]e^∞[/itex]

Last edited:

- #9

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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)`

- #10

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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.sorry. well even if that is the case:

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

then

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

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