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

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

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

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

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
$$f(x)=\frac{e^{x+\sqrt{x}}-1}{e^{\sqrt{x}}+1 e^{\sqrt{x}}-1}.$$
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: $$\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}$$ exists.

then $$\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}$$

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? :)