# Math Help - L'Hospital's Rule

1. ## L'Hospital's Rule

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}

2. Originally Posted by zachb
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}
Derivative,
(ax^{a-1}-a)/(2*(x-1))
Take derivative again,
(a(a-1)x^{a-2})/(2)
Now substitute,
(a)(a-1)/2

3. Originally Posted by zachb
2. lim x --> 0+ {sqrt(x) lnx}
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

4. Originally Posted by zachb
6. lim x --> infinity {x - lnx}
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.

5. Originally Posted by zachb
I need help finding the limits of these expression using L'Hospital's rule.

3. lim x --> 0 {cot2x sin6x}

lim x --> 0 {cot2x sin6x}

= lim x --> 0 { sin6x / tan2x} to make it 0/0 form

= lim x --> 0 { 6cos6x / 2(sec2x)^2}

= 6/2 = 3

6. lim x --> 0 {cot2x sin6x}

= lim x --> 0 { sin6x / tan2x} to make it 0/0 form

= lim x --> 0 { 6cos6x / 2(sec2x)^2}

= 6/2 = 3
how did you get from lim x --> 0 {cot2x sin6x} to lim x --> 0 { sin6x / tan2x} ?

7. Originally Posted by bobnine
how did you get from lim x --> 0 {cot2x sin6x} to lim x --> 0 { sin6x / tan2x} ?
cot(2x) = 1/tan(2x)

8. Originally Posted by zachb

4. lim x --> infinity {x^3e^-x^2}
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