Calculating limit [ Exponential ] lim [x->0] [e^(1/x) - 1]/[e^(1/x) + 1]

[Orcus]

So there is this question in my text book which is rather more puzzling for me than I would like it to be.

The question is->

Show that
Lt x->0

e^1/x-1 / e^1/x+1



Now, The question is e^1/h is infinity or undefined as h is 0 afaik. And any -1 or +1 to undefined remains undefined. However, In my book the answer has been give as -1(For LHL) and +1(For RHL). HOW?

 

[ksdhart2]

Well, my first comment is a kind of pedantic one, but it's important nonetheless. The problem statement you wrote is as follows:

\(\displaystyle \displaystyle \lim _{x\to 0}\left(e^{\frac{1}{x}}-\frac{1}{e^{\frac{1}{x}}}+1\right)\)

From context, it's easy to tell that you actually meant (e^1/x-1)/(e^1/x+1) or :

\(\displaystyle \displaystyle \lim _{x\to 0}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)\)

As you start dealing with more and more complex expressions, it may not always be easy for others to tell from context. Grouping symbols are very important.

That mini-lecture aside, you're correct that e^1/x is undefined at the point x = 0, but the limit still might exist. Consider the function [(x-2)(x+3)]/[(x-3)(x-2)]. At the point x = 2, it's undefined, but the limit as x approaches 2 is -5. To work out what the limit of e^1/x as x approaches 0 might be, let's assume the limit exists. We don't know it's value, so we'll call it L:

\(\displaystyle \displaystyle \lim _{x\to 0}\left(e^{\frac{1}{x}}\right)=L\)

Now that we have an equation, what if we take the natural log of both sides?

\(\displaystyle \displaystyle \lim _{x\to 0}\left(ln \left(e^{\frac{1}{x}}\right) \right)=ln\left(L\right)\)

Try continuing from here and see what you get. Once you know that limit, you can rewrite the problem you were given, using the "multiplication rule" and the "addition rule." (Your book/class might call them something different):

\(\displaystyle \displaystyle \lim _{x\to 0}\left(\frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}\right)=\frac{\lim _{x\to 0}\left(e^{\frac{1}{x}}\right)-\lim _{x\to 0}\left(1\right)}{\lim _{x\to 0}\left(e^{\frac{1}{x}}\right)+\lim _{x\to 0}\left(1\right)}\)

 

[Steven G]

So there is this question in my text book which is rather more puzzling for me than I would like it to be.

The question is->

Show that
Lt x->0

e^1/x-1 / e^1/x+1



Now, The question is e^1/h is infinity or undefined as h is 0 afaik. And any -1 or +1 to undefined remains undefined. However, In my book the answer has been give as -1(For LHL) and +1(For RHL). HOW?

Suppose as x->0 that x is always positive. Then 1/x goes to infinity and so e^1/x also goes to infinity.
Then adding or subtracting 1 to e^1/x is insignificant. So lim as x->0⁺ (e^1/x -1)/(e^1/x +1)= lim as x->0⁺ (e^1/x )/(e^1/x) which clearly is 1.

Now suppose as x->0 that x is always negative. Then 1/x goes to neg infinity and e^1/x goes to 0.
Then lim as x->0^- (e^1/x -1)/(e^1/x +1)= lim as x->0⁺ (-1 )/(1) which clearly is -1.

 

[HallsofIvy]

Or, since it is the 1/x that bothers you, let y= 1/x. As x goes to 0, 1/x goes to infinity so the problem becomes \(\displaystyle \lim_{y\to\infty} \frac{e^y- 1}{e^y+ 1}\). Of course, y cannot be infinity but we can take this limit.

As a general rule, when you have something that "goes to infinity", try to get it in the denominator of a fraction so that fraction goes to 0:

Divide both numerator and denominator by \(\displaystyle e^y\). Then we have \(\displaystyle \lim_{y\to\infty}\frac{1- \frac{1}{e^y}}{1+ \frac{1}{e^y}}\). Now, it is easy. As y goes to infinity so does \(\displaystyle e^y\) but that means \(\displaystyle \frac{1}{e^y}\) goes to 0. That gives \(\displaystyle \lim_{y\to 0}\frac{1- e^y}{1+ e^y}= \lim_{x\to 0}\frac{1- e^{1/x}}{1+ e^{1/x}}= \frac{1- 0}{1+ 0}= 1\).

 

[lookagain]

Or, since it is the 1/x that bothers you, let y= 1/x. As x goes to 0, 1/x goes to infinity so the problem becomes \(\displaystyle \displaystyle\lim_{y\to\infty} \frac{e^y- 1}{e^y+ 1}\). Of course, y cannot be infinity but we can take this limit.

As a general rule, when you have something that "goes to infinity", try to get it in the denominator of a fraction
so that fraction goes to 0:

Divide both numerator and denominator by \(\displaystyle e^y\). Then we have \(\displaystyle \displaystyle\lim_{y\to\infty}\frac{1- \frac{1}{e^y}}{1+ \frac{1}{e^y}}\).
As y goes to positive infinity, so does \(\displaystyle e^y,\) but that means \(\displaystyle \frac{1}{e^y}\) goes to 0.
That gives \(\displaystyle \ \ \displaystyle\lim_{y\to 0}\frac{1- e^y}{1+ e^y}= \displaystyle\lim_{x\to 0}\frac{1- e^{1/x}}{1+ e^{1/x}}= \frac{1- 0}{1+ 0}= 1\).

- - - There are two sides to the limit. - - -

Or, since it is the 1/x that bothers you, let y = 1/x. As x goes to 0 from the right, 1/x goes to positive infinity,
so the problem becomes \(\displaystyle \displaystyle\lim_{y\to\infty} \frac{e^y - 1}{e^y+ 1}\).

As a general rule, when you have something that "goes to infinity", try to get it in the denominator of a fraction so that fraction goes to 0:

Divide both numerator and denominator by \(\displaystyle e^y\). Then we have \(\displaystyle \displaystyle\lim_{y\to\infty}\frac{1- \frac{1}{e^y}}{1+ \frac{1}{e^y}}\). \(\displaystyle \ \ \)
As y goes to positive infinity, so does \(\displaystyle e^y,\) but that means \(\displaystyle \frac{1}{e^y}\) goes to 0. That gives \(\displaystyle \displaystyle\lim_{y\to \infty}\frac{1- e^y}{1+ e^y}= \displaystyle\lim_{x\to 0^+}\frac{1- e^{1/x}}{1+ e^{1/x}}= \frac{1- 0}{1+ 0}= 1.\)

- - - - - - - - - - - -


Or, since it is the 1/x that bothers you, let y = 1/x. As x goes to 0 from the left,
1/x goes to negative infinity, so the problem becomes \(\displaystyle \displaystyle\lim_{y\to - \infty} \frac{e^y - 1}{e^y + 1}\).


As y goes to negative infinity, \(\displaystyle \ e^y \) goes to 0. That gives \(\displaystyle \displaystyle\lim_{y\to -\infty}\frac{e^y - 1}{e^y + 1}= \displaystyle\lim_{x\to 0^-}\frac{e^{1/x} - 1}{e^{1/x} + 1}= \frac{0 - 1}{0 + 1}= -1.\)