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## Homework Statement

[tex]f(x)=\frac{e^{x}-e^{-\sqrt{x}}}{e^{\sqrt{x}}-e^{-\sqrt{x}}}[/tex]

show f(0)=1/2

## Homework Equations

## The Attempt at a Solution

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- Thread starter robertdeniro
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- #1

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[tex]f(x)=\frac{e^{x}-e^{-\sqrt{x}}}{e^{\sqrt{x}}-e^{-\sqrt{x}}}[/tex]

show f(0)=1/2

Last edited:

- #2

fzero

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[tex] e^{\sqrt{x}} = A e^x[/tex]

you can take a logarithm of both sides to get a more manageble equation to solve.

- #3

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[tex] e^{\sqrt{x}} = A e^x[/tex]

you can take a logarithm of both sides to get a more manageble equation to solve.

hey, sorry its fixed now

- #4

fzero

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You added "Show f(0) = 1/2" but you haven't defined f(x).

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You added "Show f(0) = 1/2" but you haven't defined f(x).

fixed that too

- #6

fzero

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fixed that too

OK, the image probably didn't reload in my browser.

Have you tried taking the limit as [tex]x\rightarrow 0[/tex]? You should really show some of your work before asking for help.

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OK, the image probably didn't reload in my browser.

Have you tried taking the limit as [tex]x\rightarrow 0[/tex]? You should really show some of your work before asking for help.

yes i tried lopital twice, didnt work and it was getting really messy so i stopped

- #8

fzero

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[tex]

f(x)=\frac{e^{x+\sqrt{x}}-1}{e^{2\sqrt{x}}-1}.

[/tex]

Then you can factor the denominator and you get an expression where you can use L'Hopital on an indeterminant factor.

- #9

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[tex]

f(x)=\frac{e^{x+\sqrt{x}}-1}{e^{2\sqrt{x}}-1}.

[/tex]

Then you can factor the denominator and you get an expression where you can use L'Hopital on an indeterminant factor.

sorry im not following

if i break up the denominator i get

[tex]

f(x)=\frac{e^{x+\sqrt{x}}-1}{e^{\sqrt{x}}+1 e^{\sqrt{x}}-1}.

[/tex]

then what?

- #10

fzero

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ok i see

but we evaluate its limit going to 0, but not AT 0.

i do not recall a theorem that says limit=actual value

- #12

fzero

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ok i see

but we evaluate its limit going to 0, but not AT 0.

i do not recall a theorem that says limit=actual value

If the function is undefined at a point, then we should attempt to define it by it's limit at that point, if such limit exists.

- #13

VietDao29

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## Homework Statement

[tex]f(x)=\frac{e^{x}-e^{-\sqrt{x}}}{e^{\sqrt{x}}-e^{-\sqrt{x}}}[/tex]

show f(0)=1/2

## Homework Equations

## The Attempt at a Solution

There are 3 ways you can go about solving this problems:

If we have:

- [tex]\lim_{x \rightarrow c} \frac{f(x)}{g(x)}[/tex] is of one of the 2
*Indeterminate Form*s [tex]\frac{0}{0}[/tex], or [tex]\frac{\infty}{\infty}[/tex]. - And the limit: [tex]\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}[/tex] exists.

You can try to apply this rule here to see if you get the desired result. :)

[tex]e ^ x = 1 + x + \frac{x ^ 2}{2!} + ...[/tex]

[tex]\lim_{x \rightarrow 0} \frac{e ^ x - 1}{x} = 1[/tex]

Remember that, this limit is

It goes like this:

[tex]\lim_{x \rightarrow 0} \frac{e ^ x - e ^ {-\sqrt{x}}}{e ^ {\sqrt{x}} - e ^ {-\sqrt{x}}} = \lim_{x \rightarrow 0} \left( \frac{e ^ {-\sqrt{x}}}{e ^ {-\sqrt{x}}} \times \frac{e ^ {x + \sqrt{x}} - 1}{e ^ {2 \sqrt{x}} - 1} \right)[/tex]

[tex]= \lim_{x \rightarrow 0} \left( \frac{e ^ {x + \sqrt{x}} - 1}{x + \sqrt{x}} \times \frac{2 \sqrt{x}}{e ^ {2 \sqrt{x}} - 1} \times \frac{x + \sqrt{x}}{2 \sqrt{x}} \right) = ...[/tex]

Can you go from here? :)

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