• Support PF! Buy your school textbooks, materials and every day products Here!

Limits probably involving e^x-1/x special limit

  • Thread starter alingy1
  • Start date
  • #1
325
0

Homework Statement


lim of (a^x-a^-x-2)/(x^2) as x->0

Find the value of the limit^.


The answer is (lna)^2
I know how to get the answer using L'Hospital.
However, I do not want to use Hospital's rule.
I think I see the pattern with the special limit:
lim of (e^x-1)/x as x->0

But I just don't see how I can apply that. I tried to take ln() out of everything, but it leads me to nowhere.
Please help me.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


lim of (a^x-a^-x-2)/(x^2) as x->0

Find the value of the limit^.


The answer is (lna)^2
I know how to get the answer using L'Hospital.
However, I do not want to use Hospital's rule.
I think I see the pattern with the special limit:
lim of (e^x-1)/x as x->0

But I just don't see how I can apply that. I tried to take ln() out of everything, but it leads me to nowhere.
Please help me.
a^x=e^(ln(a)x) is the connection between powers of e and a. You could also substitute the series expansions of the function. But I don't think knowing the special limit you've got there will help.
 
  • #3
325
0
Then, how else could we solve this problem?
My school hasn't taught me Hospital's rule. So, there must be another way.
 
  • #4
6
0
As far as I can see, L'Hopital's rule doesn't even apply since the numerator does not equal 0 if you plug in 0. a^0 - a^(-0) - 2 = 1 - 1 - 2 = -2.

You're simply dividing by a small positive number, so the limit is negative infinity.
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,187
I think there's a typo in the original post. The problem should be
$$\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}.$$ You can rewrite this as
$$\lim_{x \to 0} \frac{a^{-x}[(a^x)^2 - 2a^x +1]}{x^2}.$$ You can do a little algebra, use what Dick said about expressing this in terms of ##e##, and change variables appropriately to relate it back to the special limit.
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
I think there's a typo in the original post. The problem should be
$$\lim_{x \to 0} \frac{a^x + a^{-x} - 2}{x^2}.$$ You can rewrite this as
$$\lim_{x \to 0} \frac{a^{-x}[(a^x)^2 - 2a^x +1]}{x^2}.$$ You can do a little algebra, use what Dick said about expressing this in terms of ##e##, and change variables appropriately to relate it back to the special limit.
Ok, that's nice. Now I see how to use the 'special limit'.
 

Related Threads for: Limits probably involving e^x-1/x special limit

Replies
6
Views
11K
Replies
2
Views
964
Replies
1
Views
10K
Replies
3
Views
829
Replies
3
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
10K
  • Last Post
Replies
3
Views
3K
Top