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dan38
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Homework Statement
Don't understand why the limit (e^x - 1)/x^2 is undefined as x approaches 0?
Can't you use L'hopital's rule to get the value as 1/2?
Don't understand why the limit (e^x - 1)/x^2 is undefined as x approaches 0?
Can't you use L'hopital's rule to get the value as 1/2?
micromass said:Indeed, l'Hopitals rule only applies to situations like
[tex]\frac{0}{0}~\text{and}~\frac{\pm \infty}{\pm \infty}[/tex]
As x goes to infinity, the numerator goes to infinity and e-x goes to 0. micromass had just told you that you cannot use L'Hopitals rule for that. And you don't need to. The numerator of a fraction going to infinity while the denominator goes to 0, the fraction goes to infinity. Actually, that should have been obvious from the start. [itex]x^2e^x[/itex] is the product of two functions both of which go to infinity.dan38 said:oh i see, thanks!
doing a question: x^2*e^x with x---> infinity
changed it to
x^2/e^(-x)
Using L'hopital's rule
2x/-e^(-x)
and again
2/e^(-x)
where do I go from here? :S
2x/-e^(-x)
so how would I go from here then?
An undefined limit is a limit where the value of the function approaches either positive or negative infinity, or the function has a discontinuity at the given point. This means that the limit does not exist and cannot be determined.
The function e^x-1/x^2 is an exponential function with a polynomial in the denominator. It is commonly used in calculus to study the behavior of functions at a given point, especially when the limit approaches 0.
To find the limit of e^x-1/x^2 as x approaches 0, you can use the L'Hôpital's rule or the algebraic limit theorem. The L'Hôpital's rule states that if the limit of a quotient of two functions exists, then the limit of the quotient of their derivatives exists and is equal to the limit of the original quotient. The algebraic limit theorem states that if the limit of a quotient of two functions exists, then the limit of the product of the functions exists and is equal to the product of their limits.
The value of the limit of e^x-1/x^2 as x approaches 0 is undefined. This is because the function has a discontinuity at x=0, with the denominator approaching 0 and the numerator approaching 1. This leads to a value that approaches positive or negative infinity, meaning the limit does not exist.
The function e^x-1/x^2 is important in calculus because it is used to study the behavior of functions at a given point, especially when the limit approaches 0. It also appears in many applications, such as in the study of growth and decay, optimization problems, and differential equations. Understanding the behavior of this function can help in solving a variety of mathematical problems and real-world situations.