What is the difference in this limit?

[sylent33]

Homework Statement

Calculate the limit of the function for + and - infinity

Relevant Equations

L'Hospital

Hello!

I need to calculate the limit of this function $f(x) = (x^2-9)*e^{-x}$ for + and - $\infty$ Now for + infinity I did this

$$ \frac{(x^2-9)}{e^x} $$ apply L'Hospital since we have infinity divided by infinity; $$\frac{2x}{e^x} $$ Apply L'Hospital again $$ \frac{2}{e^x} $$ the limit is 0. Now for -$\infty$ I did exactly the same,getting 0 as my limit but apperently that is wrong.The answer should be $\infty$ and I don't know how.Intuitively that just does not make sense to me because we are simply approching it from a diffrent side but the same value.Some insight would be great,thanks!

 

[Office_Shredder]

The initial limit is not of the form infinity over infinity at minus infinity. What does $e^x$ go to?

 

[Mark44]

The initial limit is not of the form infinity over infinity at minus infinity.

@sylent33, this means that L'Hopital's Rule cannot be applied for this limit, assuming that's what you did.

 

[sylent33]

Uh okay you are right,if we have minus infinity than the $e^x$ is not going to infinity but rather to 0,so than we have infinity/0. Now I think that is 0,so I don't need to to use the l'hospital's rule but,this raises another interesting question.So l'hospital's rule says we need to get the derivative of both the nummerator and determinator.So let's assume I have a situation infinity/0, can I still apply the L'Hopital rule here?

 

[Office_Shredder]

Infinity/0 is either positive or negative infinity, depending on the sign of the denominator (or undefined if the denominator keeps flipping signs).L'hopital's rule says you need it to be 0/0 or infinity/infinity in order to take both derivatives. You can't just apply it to arbitrary limits.

 

[sylent33]

Infinity/0 is either positive or negative infinity, depending on the sign of the denominator (or undefined if the denominator keeps flipping signs).L'hopital's rule says you need it to be 0/0 or infinity/infinity in order to take both derivatives. You can't just apply it to arbitrary limits.

You are right of course infinity/0 is infinity...Making really really silly mistakes.Now since the concept of infinity is very abstract and I know that we cannot consider it a number,but ocassionaly the situation comes in let's say calculating limits of a sequence or a function where we have $2 *\infty$ or $2 * -\infty$ and most of these are usually though in class.But not to long ago I stumbled upon an interesting situation,it was a constant number(negative constant number) times - infinity.Something like this

$-2 * \infty$ and the result was $-\infty$.Now I know that we cannot consider infinity to be a number,so I am curious as to how does one approach this,is there a certain method you have to use to evalue this,or is it something you would find on a formula sheet?

 

[Mark44]

You are right of course infinity/0 is infinity...

Not necessarily. If the denominator is approaching zero through the negative numbers, the limit won't be infinity.

Making really really silly mistakes.Now since the concept of infinity is very abstract and I know that we cannot consider it a number,but ocassionaly the situation comes in let's say calculating limits of a sequence or a function where we have $2 *\infty$ or $2 * -\infty$ and most of these are usually though in class.But not to long ago I stumbled upon an interesting situation,it was a constant number(negative constant number) times - infinity.Something like this

$-2 * \infty$ and the result was $-\infty$.Now I know that we cannot consider infinity to be a number,so I am curious as to how does one approach this,is there a certain method you have to use to evalue this,or is it something you would find on a formula sheet?

If you have a function whose limit is infinity, what this really means is that the function values get arbitrarily large as the independent variable approaches its limiting value.

So if you're given that $\lim_{x \to a} f(x) = \infty$, then the following will be true.
$c\lim_{x \to a} f(x) = \infty$, if c is finite and strictly positive.
$c\lim_{x \to a} f(x) = -\infty$, if c is finite and strictly negative.
If c = 0, you can't say what the limit will be (if any) without doing some more work.

Also, if $\lim_{x \to a} g(x) = \infty$, then the following will be true.
$\lim_{x \to a} \left(f(x) + g(x)\right) = \infty$.
But $\lim_{x \to a} \left(f(x) - g(x)\right)$ is indeterminate, as is $\lim_{x \to a} \frac{f(x)}{g(x)}$.

 

[sylent33]

Not necessarily. If the denominator is approaching zero through the negative numbers, the limit won't be infinity.
If you have a function whose limit is infinity, what this really means is that the function values get arbitrarily large as the independent variable approaches its limiting value.

So if you're given that $\lim_{x \to a} f(x) = \infty$, then the following will be true.
$c\lim_{x \to a} f(x) = \infty$, if c is finite and strictly positive.
$c\lim_{x \to a} f(x) = -\infty$, if c is finite and strictly negative.
If c = 0, you can't say what the limit will be (if any) without doing some more work.

Also, if $\lim_{x \to a} g(x) = \infty$, then the following will be true.
$\lim_{x \to a} \left(f(x) + g(x)\right) = \infty$.
But $\lim_{x \to a} \left(f(x) - g(x)\right)$ is indeterminate, as is $\lim_{x \to a} \frac{f(x)}{g(x)}$.

Great answer but I am not sure I get the "if the denominator part is approaching zero through the negative numbers".But than it wouldn't even be infinity/0, but rather infinity/-infinity?

 

[Office_Shredder]

Consider $\lim_{x\to \infty} 1/(e^{-x})$ vs $\lim_{x\to \infty} 1/(-e^{-x})$ In both cases it goes to 1/0, but one is infinity and the other is negative infinity. The sign of 0 is not determined.

 

[Mark44]

Great answer but I am not sure I get the "if the denominator part is approaching zero through the negative numbers".But than it wouldn't even be infinity/0, but rather infinity/-infinity?

No, it wouldn't be of the form infinity/-infinity if the denominator is getting close to zero (but is always negative).

 

[sylent33]

I think I get it now,shredder gave an great example