Vtyjkyuk

I want to evaluate the following limit:

$$lim_xrightarrow0frace^x-e^-xe^x-1$$

For $xrightarrow0$, the denominator is asymptotic to

$$e^x-1sim x$$

Here's how I simplify the numerator:

$$e^x-e^-xsim 1-e^-x=-(e^-x-1)sim-(-x)=x$$

Finally we have

$$f(x)simfracxx=1$$

My textbook does it another way and the solution is 2. I wonder what I'm doing wrong. Perhaps it's the fact that I have replaced $e^x$ with $1$ but that seemed to make sense to me because $e^xrightarrow1$ for $xrightarrow0$. Any hints?

How do you justify $e^x-e^-xsim 1-e^-x$?

It is true that $e^x = 1 + o(1)$, but if that is your justification then $sim$ must mean "up to terms of degree $1$ or more", so when you end up with the numerator being $sim x$, then this is the same as $sim 0$ or $sim 2x$ as far as this meaning of $sim$ is concerned. You've thrown away the information you need to conclude anything.

What you did is wrong because both $e^x$ and $e^-x$ behave as $1$ near $0$. You can't replace one if them by $1$, while the other one remains as it is.

Note that$$lim_xto0frace^x-e^-xe^x-1=lim_xto0frace^x+e^-xe^x=2.$$

The solution goes as follows:
$$lim_xto 0frace^x-e^-xe^x-1=lim_xto 0fracfrace^2x-1e^xe^x-1=lim_xto 0fracfrac(e^x-1)(e^x+1)e^xe^x-1=lim_xto 0frace^x+1e^x=2$$

$displaystylelim_x to 0dfrace^x-e^-xe^x-1$

now,
$e^x-e^-x=left(1+x+dfracx^22!+dfracx^33!+cdots right)-left(1-x+dfracx^22!-dfracx^33!+cdots right)=2left(x+dfracx^33!+dfracx^55!+cdots right) approx2x$ as $ xto 0$

and $e^x-1=x+dfracx^22!+dfracx^33!+cdots approx x $, as $ xto 0$

$therefore displaystylelim_x to 0dfrace^x-e^-xe^x-1=displaystylelim_x to 0dfrac2xx=2.$

You cannot just substitute the limits, as that leads to $0/0$. L'Hopital's rule is designed for exactly this type of problems, see:
https://en.wikipedia.org/wiki/L%27Hôpital%27s_rule

And indeed, the answer is $2/1=2$, once you apply that rule.

Using equivalents as you did (which the simplest method here), observe the numerator is
$$mathrm e^x-mathrm e^-x=2sinh xsim_0 2x,$$
whence
$$fracmathrm e^x-mathrm e^-xmathrm e^x-1sim_0frac2xx=2.$$

Making $y = e^x$ we have

$$lim_xrightarrow0frace^x-e^-xe^x-1equiv lim_yto 1fracy-frac 1yy-1 = lim_yto 1fracy^2-1y(y-1) = lim_yto 1fracy+1y= 2 $$