How to evaluate this function?

[robertdeniro]

Homework Statement

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

show f(0)=1/2

Homework Equations

The Attempt at a Solution

 

[fzero]

I think you have a sign wrong somewhere, otherwise the LHS = 1. But if you can express your equation in the form

e^{\sqrt{x}} = A e^x

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

 

[robertdeniro]

I think you have a sign wrong somewhere, otherwise the LHS = 1. But if you can express your equation in the form

e^{\sqrt{x}} = A e^x

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

hey, sorry its fixed now

 

[fzero]

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

 

[robertdeniro]

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

fixed that too

 

[fzero]

fixed that too

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

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

 

[robertdeniro]

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

Have you tried taking the limit as x\rightarrow 0? 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

 

[fzero]

I think it helps to note that you can rewrite

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

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

 

[robertdeniro]

I think it helps to note that you can rewrite

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

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

sorry I am not following

if i break up the denominator i get
<br /> f(x)=\frac{e^{x+\sqrt{x}}-1}{e^{\sqrt{x}}+1 e^{\sqrt{x}}-1}.<br />
then what?

 

[fzero]

Write that as a product of two factors, one of which is well-defined in the limit and the other which is indeterminate. L'Hopital can be applied to the indeterminate factor.

 

[robertdeniro]

Write that as a product of two factors, one of which is well-defined in the limit and the other which is indeterminate. L'Hopital can be applied to the indeterminate factor.

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

 

[fzero]

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.

 

[VietDao29]

Homework Statement

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

show f(0)=1/2

Homework Equations

The Attempt at a Solution

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

1. The first way is to use L'Hopital's Rule: Have you studied L'Hopital's Rule yet? The L'Hopital's Rule states that:

If we have:

• \lim_{x \rightarrow c} \frac{f(x)}{g(x)} is of one of the 2 Indeterminate Forms \frac{0}{0}, or \frac{\infty}{\infty}.
• And the limit: \lim_{x \rightarrow c} \frac{f&#039;(x)}{g&#039;(x)} exists.

then \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f&#039;(x)}{g&#039;(x)}

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

2. The second way is to use Taylor expansion.
e ^ x = 1 + x + \frac{x ^ 2}{2!} + ...

3. The last way, also the most fundamental way is to use the well-known limit:
\lim_{x \rightarrow 0} \frac{e ^ x - 1}{x} = 1

Remember that, this limit is very useful when dealing with problems which ask you to take the limit of exponential function!

It goes like this:
\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)
= \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) = ...

Can you go from here? :)