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How to evaluate this function?

  • #1

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

 
Last edited:

Answers and Replies

  • #2
fzero
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I think you have a sign wrong somewhere, otherwise the LHS = 1. But if you can express your equation in the form

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

[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).
 
  • #5
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.
 
  • #7
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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I think it helps to note that you can rewrite

[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
I think it helps to note that you can rewrite

[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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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.
 
  • #11
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
 
  • #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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2

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:

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:
  • [tex]\lim_{x \rightarrow c} \frac{f(x)}{g(x)}[/tex] is of one of the 2 Indeterminate Forms [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.
then [tex]\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}[/tex]

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

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

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:
[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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