I need help finding the limits of these expression using L'Hospital's rule.
1. lim x --> 1 {x^a - ax + a -1/(x -1)^2}
2. lim x --> 0+ {sqrt(x) lnx}
3. lim x --> 0 {cot2x sin6x}
4. lim x --> infinity {x^3e^-x^2}
5. lim --> infinity {xe^1/x - x}
6. lim --> infinity {x - lnx}
5. lim --> 0+ {x^x^2}
This is in the form 0*(infinity), so do the following:
lim(x-->0+){sqrt(x) lnx} = lim(x -->0+){ln(x)/[1/sqrt(x)]}
Now we have an (infinity)/(infinity) format for the limit, so we can take L'Hopital's rule:
lim(x-->0+){sqrt(x) lnx} = lim(x-->0+){[1/x]/[(-1/2)/sqrt(x^3)]}
= (-2)lim(x-->0+){sqrt(x^3)/x}
= (-2)lim(x-->0+){sqrt(x)} = 0
-Dan
Think of x as ln(e^x). Then
x - ln(x) = ln(e^x) - ln(x) = ln[e^x/x]
Now, if lim(x-->infinity){e^x/x} is finite, then lim(x-->infinity){ln[e^x/x]} is finite. (If lim(x-->infinity){e^x/x} = 0 then lim(x-->infinity){ln[e^x/x]} is negative infinity.) If lim(x-->infinity){e^x/x} is infinite, then lim(x-->infinity){ln[e^x/x]} is infinite.
So let's take a look at: lim(x-->infinity){e^x/x}. This is of the form (infinity)/(infinity) so use L'Hopital's rule:
lim(x-->infinity){e^x/x} = lim(x-->infinity){e^x/1} = (infinity)
Thus
lim(x -->infinity){x - lnx} --> (infinity)
-Dan
PS Note that I would need to be VERY careful with this argument if lim(x-->infinity){e^x/x} had been finite. In general it is not a good idea to switch the order of the limit and ln operators (or the limit and any other operator.) As the limit turned out to be infinite this argument didn't matter.
lim x --> infinity {x^3e^-x^2}
= lim x --> infinity {x^3/e^x^2} ......infinity/infinity
Apply L'Hopital's
= lim x --> infinity {3x^2/2x(e^x^2)}
= lim x --> infinity {(3/2) x/(e^x^2)} ........(3/2)*infinity/infinity
Apply L'Hopital's again
= lim x --> infinity {(3/2)1/2x(e^x^2)}
= 0