Uni12345678
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I have a set of limits questions, which I’m really confused with!!

1. Why does (X^2)*e^X go to infinity as x tends to infinity ? Do we have to use l’hopital’s ?

2. Why does cos(x)/(x-pi/2)^2 go to infinity at pi/2? I thought it would be indeterminate and then it would go to 0 using l’hopitals

3. Why does (x^2 + 2x)/x^5 go to infinity as x tends to 0? I thought using l’hopitals we’d end up with 6x leading to 0 as the limit.

Any help on these 3 would be much appreciated!!
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RDKGames
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(Original post by Uni12345678)
I have a set of limits questions, which I’m really confused with!!

1. Why does (X^2)*e^X go to infinity as x tends to infinity ? Do we have to use l’hopital’s ?

2. Why does cos(x)/(x-pi/2)^2 go to infinity at pi/2? I thought it would be indeterminate and then it would go to 0 using l’hopitals

3. Why does (x^2 + 2x)/x^5 go to infinity as x tends to 0? I thought using l’hopitals we’d end up with 6x leading to 0 as the limit.

Any help on these 3 would be much appreciated!!
1. No need for L'Hopitals. I'd say this should be very obvious. Both terms in the product, x^2 and e^x, individually tends to \infty so of course the product will as well.

2. You can apply L'Hopital's rule here once and see that this limit is equivalent to \dfrac{-\sin x}{2(x-\pi/2)} as x\to \pi/2. The numerator approaches a finite value but the denominator tending to zero makes the whole limit shoot off to infinity.

3. One application of L'Hopital's lands you on \displaystyle \lim_{x \to 0} \dfrac{2x + 2}{5x^4}. The numerator tends to the finite value of 2, but the denominator going to zero makes the whole limit shoot off to infinity.
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Uni12345678
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(Original post by RDKGames)
1. No need for L'Hopitals. I'd say this should be very obvious. Both terms in the product, x^2 and e^x, individually tends to \infty so of course the product will as well.

2. You can apply L'Hopital's rule here once and see that this limit is equivalent to \dfrac{-\sin x}{2(x-\pi/2)} as x\to \pi/2. The numerator approaches a finite value but the denominator tending to zero makes the whole limit shoot off to infinity.

3. One application of L'Hopital's lands you on \displaystyle \lim_{x \to 0} \dfrac{2x + 2}{5x^4}. The numerator tends to the finite value of 2, but the denominator going to zero makes the whole limit shoot off to infinity.
This is really useful thank you

I think I’m mainly getting confused with the idea of the denominator being 0- I thought this meant it’s indeterminate so I just did the l’hopitals again and again

How do you tell whether it’s diverging or whether to do l’hopitals again?
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Uni12345678
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This is really useful thank you

I think I’m mainly getting confused with the idea of the denominator being 0- I thought this meant it’s indeterminate so I just did the l’hopitals again and again

How do you tell whether it’s diverging or whether to do l’hopitals again?
And one other tiny question, for the first question, if x tends to minus infinity, what happens ?
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RDKGames
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(Original post by Uni12345678)
This is really useful thank you

I think I’m mainly getting confused with the idea of the denominator being 0- I thought this meant it’s indeterminate so I just did the l’hopitals again and again

How do you tell whether it’s diverging or whether to do l’hopitals again?
You should check what an indeterminate form looks like again. After one application of L'Hopital you lose the indeterminate form so you cannot apply L'Hopital again.

That last question doesn't make much sense. You can tell whether a quotient is diverging or not by seeing whether either the numerator grows to infinity faster than the denominator, and/or the denominator itself tends to zero.
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RDKGames
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(Original post by Uni12345678)
And one other tiny question, for the first question, if x tends to minus infinity, what happens ?
The product goes to zero. Simply because the exponential decays faster than the quadratic grows, hence it dominates the behaviour of the limit.

Or if you want to use L'Hopitals, rewrite this as \displaystyle \lim_{x\to-\infty} \dfrac{x^2}{e^{-x}} whereby both the numerator and denominator go off to infinity giving you an indeterminate form to apply L'Hopital to.
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Uni12345678
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(Original post by RDKGames)
You should check what an indeterminate form looks like again. After one application of L'Hopital you lose the indeterminate form so you cannot apply L'Hopital again.

That last question doesn't make much sense. You can tell whether a quotient is diverging or not by seeing whether either the numerator grows to infinity faster than the denominator, and/or the denominator itself tends to zero.
So does this mean if it’s going to infinity, it’s essentially diverging? So would question 2 and 3 be diverging at those values?

I understand the point about l’hopitals now- so once one value is finite, it’s no longer indeterminate - have I got this right?

I’m also wondering what happens if in the first question x tends to minus infinity- using l’hopitals ?
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RDKGames
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(Original post by Uni12345678)
So does this mean if it’s going to infinity, it’s essentially diverging? So would question 2 and 3 be diverging at those values?

I understand the point about l’hopitals now- so once one value is finite, it’s no longer indeterminate - have I got this right?

I’m also wondering what happens if in the first question x tends to minus infinity- using l’hopitals ?
Yes; diverging means going off to infinity.

We say the limit does not exist for questions 2 and 3, because it diverges to infinity.

As for your Lhopital's point, I suppose so? You should really just look at the conditions where it is applicable. https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

It's effectively when you have a quotient and both numerator and denominator go to zero, or \pm \infty.

I answered your query concerning x \to -\infty in my last post.
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davros
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(Original post by Uni12345678)
I have a set of limits questions, which I’m really confused with!!

1. Why does (X^2)*e^X go to infinity as x tends to infinity ? Do we have to use l’hopital’s ?

2. Why does cos(x)/(x-pi/2)^2 go to infinity at pi/2? I thought it would be indeterminate and then it would go to 0 using l’hopitals

3. Why does (x^2 + 2x)/x^5 go to infinity as x tends to 0? I thought using l’hopitals we’d end up with 6x leading to 0 as the limit.

Any help on these 3 would be much appreciated!!
Are you actually meant to be using l'Hopital's rule for these? They can all be tackled more directly without using l'Hopital!
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