Sometimes we run into cases when both the denominator and numerator seem to go to infinity, and it is hard for us to find whether denominator or numerator changes at a faster rate, which means we cannot get the limit in this case. But with the help of L'Hospital's Rule, we can make this impossible calculation become possible. All we do is just to take deriatives seperately on the denominator and numerator, The limit of the ratio between numerator and denominator now becomes the limit of the ratio between derivative of numerator and derivative of denominator. If the ratio is still infinity over infinity, just take derivative again until there is the answer you are looking for.
For Example, `lim_(x->oo) (e^x)/x = lim (e^x)/1 =oo`
The derivative of the numerator e^x is e^x, and the derivative of denomenator x is 1, so lim (e^x)/1 =oo
But this is too easy and is not likely to be in exams. So let's try something harder based on this.
What about `lim_(x->oo) (e^x)/(x^100000050000)`
In this problem, we can just observe a trend from a series of calculations:
We have already got `lim_(x->oo) (e^x)/x=oo `above, we can do more since they look similar:
`lim_(x->oo) (e^x)/(x^2)=lim_(x->oo) (e^x)/(2x)=(1/2)*lim_(x->oo) (e^x)/x=(1/2)*oo=oo`We plug in the result from the previous step and that is how we get (1/2) *oo.
`lim_(x->oo) (e^x)/(x^3)=lim_(x->oo) (e^x)/(3x^2)=(1/3)*lim_(x->oo) (e^x)/(x^2)=(1/3)*oo=oo`Again we plug in the result from the second previous step and that is how we get (1/3)*oo
Now we see the trend that it always goes to infinity. so the answer we are looking for is (1/100000050000)*`oo=oo.`
Calculus symbol is really hard to type here. Hope you still understand the idea of it.