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Solutions to various limit problems using L'Hopital's rule and identification of indeterminate forms. It covers limits approaching 0, infinity, and the application of ln and exp functions.
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Worksheet 7, Math 10560 1 Find lim x→ 0 sin x − x x^3
Solution: (^) xlim→ 0 sin x − x x^3 is indeterminate of type 00. By L’Hopital’s rule it equals
xlim→ 0
(cos x) − 1 3 x^2
This is still an indeterminate form of type 00 , so we apply L’Hopital’s rule again. Our limit is now equal to lim x→ 0 − sin x 6 x
Applying L’Hopitals rule again or using the fact that limx→ 0 sinx^ x= 1, we get that the above limit equals − 61.
2 Find lim x→∞
x^2 ex^
(a) 0
(b) e^2
(c) ∞
(d) (^2) e
(e) Does no exist
3 Find lim t→ 0 arcsin(t) t
(a) 1
(b) 0
(c) +∞
(d) π 6
(e) Does not exist and is not +∞
4 Example Evaluate the limit lim x→ 0 +^ x^2 ln x.
Solution: We have that
lim x→ 0 +^ x^2 ln x = lim x→ 0 +
ln x x−^2 = lim x→ 0 +
x−^1 − 2 x−^3 = lim x→ 0 +^
x^2 = 0
where we used l’Hopital’s rule for the second equality. We can do that since both ln x and x−^2 go to ±∞ when x → 0 +.
7 Evaluate the following limit.
xlim→∞
(ln x)^2 x
Please show your work.
Solution: Use l’Hospital’s Rule twice:
xlim→∞
(ln x)^2 x = (^) xlim→∞
2(ln x)(1/x) 1 [l’Hospital for “∞/∞”]
= (^) xlim→∞ 2 ln x x
= (^) xlim→∞ 2(1/x) 1 [l’Hospital for “∞/∞”]
= (^) xlim→∞
x = 0.
8 Compute the limit
lim x→ 2
(x 2
) (^) x (^1) − 2 .
Solution: Solution: We have an indeterminate form 1∞. Let L = limx→ 2
(x 2
) (^) x− (^12)
. Then
ln L = lim x→ 2 ln
(x 2
) (^) x^1 − 2 = lim x→ 2
ln
(x 2
x − 2 (l’Hospital’s rule) = lim x→ 2
x
Therefore L = e (^12) =
e.
9 Evaluate the limit lim x→ 0
cos x
) 1 /x^2 .
Solution: Solution: Solution: The limit has indeterminate form 1∞. Let L = limx→ 0
cos(x)
) 1 /x^2 .
ln L = lim x→ 0 ln
cos(x)
) 1 /x 2 ) = lim x→ 0
ln
cos(x)
x^2
(Indet. form 00. Use L’Hospital’s Rule)
= lim x→ 0
1 cos(x) (−^ sin(x)) 2 x
= lim x→ 0
− tan(x) 2 x
(Indet. form 00. Use L’Hospital’s Rule)
= lim x→ 0 -sec^2 (x) 2
Therefore L = e−^ (^12) .
