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

In summary: I just need to find a way to show that$$\lim_{x \to 0} \frac{(a^x)^2 - 2a^x +1}{x^2} = (\ln a)^2.$$In summary, the conversation discusses finding the value of the limit of (a^x-a^-x-2)/(x^2) as x approaches 0. The answer is (lna)^2 and the use of L'Hospital's rule is not desired. The conversation also touches on the special limit lim of (e^x-1)/x as x approaches 0 and the connection between powers of e and a. It is suggested to rewrite the problem as lim of (a^
  • #1
alingy1
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.
 
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  • #2
alingy1 said:

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
Then, how else could we solve this problem?
My school hasn't taught me Hospital's rule. So, there must be another way.
 
  • #4
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
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
vela said:
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'.
 

What is the special limit involving e^x-1/x?

The special limit involving e^x-1/x is:
lim x->0 (e^x-1)/x = 1.

What is the significance of this special limit?

The significance of this special limit is that it is commonly used in calculus and other areas of mathematics to solve various problems involving rates of change and optimization.

How is this limit derived?

This limit can be derived using L'Hopital's rule, by taking the derivative of both the numerator and denominator with respect to x and then evaluating the limit as x approaches 0.

What is the relationship between this limit and the natural logarithm function?

The natural logarithm function, ln(x), is the inverse of the exponential function, e^x. Therefore, the special limit involving e^x-1/x can also be written as lim x->0 ln(1+x)/x = 1.

How can this limit be used to solve real-world problems?

This limit can be used to solve real-world problems involving rates of change, such as finding the maximum or minimum value of a function or calculating the velocity or acceleration of an object at a specific point in time.

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