Akward Limit: Applying l'Hospital's Rule to lim(x->-oo) 4/(2+x.e^x)

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e_2.718281828

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lim(x->-oo) 4/(2+x.e^x)
=4/(2-oo.(1/oo))
Hôpital
=0/(e^x+(e^x).x)
=0

The graph says i'm right
It's very likely that my math is wrong

I highly recommend you write down the few calculations I made, as they contain numbers divided by fractions
 
Last edited:
e_2.718281828 said:
lim(x->-oo) 4/(2+x.e^x)
=4/(2-oo.(1/oo))
Hôpital
=0/(e^x+(e^x).x)
=0

The graph says i'm right
It's very likely that my math is wrong

I highly recommend you write down the few calculations I made, as they contain numbers divided by fractions
Click to expand...

You can't directly apply l'Hôpital's rule, because the entire expression does not have the form infinity/infinity or 0/0. (Also, it appears that you didn't check whether your new denominator is also zero, in which case you can't say the limit is zero even if your application of l'Hôpital were valid.)

You might, however, use l'Hôpital to find the limit of x.e^x, and then use that result to find the desired limit.
 
Dr.Peterson said:
You can't directly apply l'Hôpital's rule, because the entire expression does not have the form infinity/infinity or 0/0. (Also, it appears that you didn't check whether your new denominator is also zero, in which case you can't say the limit is zero even if your application of l'Hôpital were valid.)

You might, however, use l'Hôpital to find the limit of x.e^x, and then use that result to find the desired limit.
Click to expand...
Can I use l'hopital's rule if it's -oo/oo ? or do both parts of the fraction have to be positive?
Thanks a lot for the advice, I hadn't thought of the problem that way
 
e_2.718281828 said:
Can I use l'hopital's rule if it's -oo/oo ? or do both parts of the fraction have to be positive?
Thanks a lot for the advice, I hadn't thought of the problem that way
Click to expand...

Signs aren't the issue; it just has to be some sort of infinity over some sort of infinity. Check the precise statement of the theorem that you were given, or one on a site like this.

If you need more help, show what you've done, either as I suggested, or rewriting the entire function so that it fits the requirements of the theorem.
 
Dr.Peterson said:
Signs aren't the issue; it just has to be some sort of infinity over some sort of infinity. Check the precise statement of the theorem that you were given, or one on a site like this.

If you need more help, show what you've done, either as I suggested, or rewriting the entire function so that it fits the requirements of the theorem.
Click to expand...
I'm fine, I found the answer(it's 2 since lim-oo xe^x=0), thanks a lot. Your advice made me remember something the teacher showed us.
 
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