Limit of e(-t/2)((k/2)t+c) as t Approaches Infinity

• wumple
In summary, the limit of e(-t/2)((k/2)t+c) as t approaches infinity where k and c are constants is 0. This can be shown by expanding the expression and observing that the term containing e(-t/2) approaches 0 faster than any polynomial can approach infinity as t approaches infinity. L'Hopital's rule can also be used to show this by taking the derivative of the expression and observing that it does not contain a term in 't'.
wumple

Homework Statement

Find the limit of e(-t/2)((k/2)t+c) as t approaches infinity where k and c are constants

Not sure..?

The Attempt at a Solution

Plugging in t = infinity gives me an indeterminate form, and multiple applications of L'hopital's rule have led me no where. Any suggestions? I can see graphically that it goes to 0, but I'm not sure how to show this analytically. I can see that if I expand it, the e(-t/2)c term goes to zero, but I'm not sure about the other term.

I do not know if this is a valid form of proving limits, but e-t/2 approaches 0 faster than any polynomial can approach infinity as t→∞.

yeah I was hoping that L'hopital's rule would show that but it didn't work out...

wumple said:
yeah I was hoping that L'hopital's rule would show that but it didn't work out...

Wouldn't L'Hopital's rule as show it going to zero since d/dt{0.5kt+c} is does not contain a tern in 't'?

rock.freak667 said:
Wouldn't L'Hopital's rule as show it going to zero since d/dt{0.5kt+c} is does not contain a tern in 't'?

Oops! yes! Thank you, I found my mistake.

1. What is the limit of e(-t/2)((k/2)t+c) as t approaches infinity?

The limit of e(-t/2)((k/2)t+c) as t approaches infinity is equal to zero. This is because as t gets larger and larger, the exponential term with a negative exponent will approach zero, making the entire expression equal to zero.

2. How do you calculate the limit of e(-t/2)((k/2)t+c) as t approaches infinity?

To calculate the limit of e(-t/2)((k/2)t+c) as t approaches infinity, we first need to determine the dominant term in the expression. In this case, as t approaches infinity, the exponential term with a negative exponent becomes the dominant term. Therefore, the limit can be calculated by plugging in infinity for t in this dominant term, resulting in a value of zero.

3. Can the limit of e(-t/2)((k/2)t+c) as t approaches infinity be negative?

No, the limit of e(-t/2)((k/2)t+c) as t approaches infinity cannot be negative. This is because the exponential term with a negative exponent will always approach zero as t gets larger, making the entire expression equal to zero. Therefore, the limit can only approach zero or be equal to zero, but it cannot be negative.

4. How does the value of k affect the limit of e(-t/2)((k/2)t+c) as t approaches infinity?

The value of k does not affect the limit of e(-t/2)((k/2)t+c) as t approaches infinity. This is because as t gets larger, the exponential term with a negative exponent becomes the dominant term, making the value of k irrelevant in determining the limit.

5. What happens to the limit of e(-t/2)((k/2)t+c) as t approaches infinity if c is a negative value?

If c is a negative value, the limit of e(-t/2)((k/2)t+c) as t approaches infinity will still be equal to zero. This is because as t gets larger, the exponential term with a negative exponent will approach zero regardless of the value of c. Therefore, the negative value of c will not affect the limit.

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