Calculus II
Practice Problems 5: Answers
1. lim
x 0
sin x x
x cos x 1
Answer. l’Hôpital’s rule will apply, and continues to apply as we use it, so we get the train of equalities
sin x x
x cos x 1
lim
x 0
lH
lim
x 0
cosx 1
cos x 1 x sinx
lH
cosx
3 cos x xsin x
since the last limit can be evaluated at the limit point.
lim
x 0
ex 1 x
0
x2
2. lim
x
lim
x 0
1
3
sin x
2 sin x xcos x
lH
Answer. Again, l’Hôpital’s rule will apply through this train of equalities:
ex 1 x
0
x2
lim
x
3. lim
x 1
ln x
cos π 2 x lH
ex 1
0 2x
lim
x
lH
ex
0 2
x
1
2
lim
Answer. After verifying that the hypotheses of l’Hôpital’s rule hold:
lim
x 1
ln x
cos π 2 x lH
lim
x 1
1 x
1
π 2 sin π 2 x π 2 1
cos x 4. lim
x
x 0
cos x 1 l H lim lim
x 0
x
x 0
since sin u u 1 as u 0.
Answer.
5 cos π x x 5
x2 25
5. lim
x
sin
Answer. We can use the same methods since cos 5π since sin 5π 0.
x 1
2 x
1
5cos π x x l H lim lim
x 5
x 5
x2 25
1:
sin x 2 x
lim
x 0
5π sin π x 1
2x
1
10
2
π
1
2
6. lim x 1 ln ln x x 1
Answer. Since x 1 We find
0 and ln ln x ∞, this is of the form 0 ∞, and we have to invert one of the factors.
lim x 1 ln ln x x 1
ln ln x 1 x 1 lim
x 1
1
lH
x ln x
1 x 1 2 lim
x
lim
x 1
x 1 2
x ln x
using a little algebra. Now, apply l’Hôpital’s rule again:
lH
7. lim
x ∞
x
1 x2
lim
x 1
2 x 1
ln x 1
0
Answer. l’Hôpital’s rule applies, and we get
∞
lim
x
x
1 x2
lH
lim
x ∞
x
Thus the number we are looking for is its own inverse, so is
positive, the limit must be 1.
We note that a little algebra will do as well:
lim
x ∞
8. lim
x 1
1
ln x
1
x 1
x
1
x2
1
1 x2
lim
x ∞
1 x2
x
lim
x ∞
1. However, since the values, as x ∞ are
1
1
x2
1
1
Answer. This is of the form ∞ ∞, so we need to start by using some algebra. Putting the expression over a
common denominator, we find
1 1 x 1 lnx
ln x x 1 x 1 ln x Now we can use l’Hôpital’s ule:
lim
x 1
x 1 lnx
x 1 lnx
lH
lim
1 1 x
lnx
x 1 x 1
x
lim
x 1
x 1
x 1 xln x
lH
lim
x 1
1
1 ln x 1
1
2