Exploring the Limit of (a^x - 1)/x as x Approaches 0

  • Thread starter wiz0r
  • Start date
  • Tags
    Limit
In summary, the limit as x approaches 0 of (a^x - 1)/x is equal to log(a). This can be proven by using L'Hospital's rule or by taking the Maclaurin series of a^x and substituting it into the limit expression. Both approaches will result in the same answer.
  • #1
wiz0r
57
0

Homework Statement



lim x->0 ( (a^x - 1)/x )

Homework Equations



NA

The Attempt at a Solution



The professor told me that the answer to that limit is log(a), but why? I don't understand; can someone explain why?
 
Physics news on Phys.org
  • #2
Do you know l'hopital's rule?
 
  • #3
gb7nash said:
Do you know l'hopital's rule?


Yeah, I just figured it out. Take the derivative of both the numerator and the denominator and then it's easy from there...

Is there any way to do it without having to use L'Hospital's?
 
  • #4
You could take the maclaurin series of a^x and plug it in for a^x. Once you do this, stuff cancels out and you'll obtain the same answer.
 

What is the limit of (a^x - 1)/x as x approaches 0?

The limit of (a^x - 1)/x as x approaches 0 is equal to ln(a), where ln is the natural logarithm function.

Why is it important to explore the limit of (a^x - 1)/x as x approaches 0?

Exploring this limit is important because it can help us understand the behavior of exponential functions and their rates of change at a specific point. It can also have applications in areas such as calculus, physics, and finance.

How do you find the limit of (a^x - 1)/x as x approaches 0?

To find the limit, we can use the L'Hopital's rule or algebraic manipulation. L'Hopital's rule states that if we have a limit of the form f(x)/g(x) as x approaches 0, and both f(x) and g(x) approach 0 or infinity, then the limit is equal to the limit of the quotient of their derivatives. Algebraic manipulation involves factoring out an x from the numerator and denominator and then simplifying the expression.

What does the value of the limit of (a^x - 1)/x as x approaches 0 tell us?

The value of this limit tells us the slope of the tangent line to the exponential function at the point where x = 0. It also tells us the rate of change of the function at that specific point.

Are there any restrictions on the value of a when exploring the limit of (a^x - 1)/x as x approaches 0?

Yes, a must be a positive real number, as the natural logarithm function is only defined for positive numbers. Additionally, a cannot be equal to 1, as the limit would then be undefined.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
652
  • Calculus and Beyond Homework Help
Replies
12
Views
769
  • Calculus and Beyond Homework Help
Replies
19
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
664
  • Calculus and Beyond Homework Help
Replies
10
Views
820
  • Calculus and Beyond Homework Help
Replies
6
Views
880
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
722
  • Calculus and Beyond Homework Help
Replies
30
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
233
Back
Top