1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
November 7
LH
Compute these limits. Note = means that L’Hopital’s rule ws used.
5 sin(x) + e2x − 1 LH
5 cos(x) + 2e2x
= lim
=5+2=7
x→0
x→0
x
1
1. lim
ex + e−x
=
x→0
x
This is not an indeterminate form since when you plug in you get
rule does not apply.
2. lim
Since lim+
x→0
1+1
0
= 20 . Thus L’Hopitals
ex + e−x
ex + e−x
ex + e−x
= ∞ and lim−
= −∞ we find that lim
=DNE
x→0
x
x
x
x→0
ex + e−x − 2 LH
ex − e−x LH
ex + e−x
2
1
= lim
= lim
= =
x→0 1 − cos(2x)
x→0 2 sin(2x)
x→0 4 cos(2x)
4
2
√
−1
1 2
√ x
2 ∗ 2x
x2 + 4 LH
x2 +4
2 (x + 4)
4. lim
=
lim
=
lim
x→∞
x→∞
x→∞ 3e3x
e3x
3e3x
3. lim
x
√
. This is another indeterminate
x→∞ 3e3x x2 + 4
form, ∞
∞ . Now we would apply L’Hopital’s Rule again. If you did this then you would notice
that the limit does not get any nicer.
Using the techniques for limits to infinity that we learned at the start of the semester, we see
x
that lim √
= 1. Combine this with the fact that 3e3x goes to infinity as x goes to infinity,
x→∞ x2 + 4
we get
We could simplify some the above limit to lim
√ x
x2 +4
x→∞ 3e3x
lim
5.
lim
x→ π2 −
=0
4 tan x LH
4 sec2 x
4 sec x
= lim
= lim
− sec(x) tan(x)
− tan(x)
1 + sec x
x→ π
x→ π
2
2
This limit is still inderminate, ∞
∞ , and applying L’Hopitals’s rule will continue a pattern of fractions with secant and tangent of being on top or on bottom of the fractions. To avoid this,
convert the trig functions to sine and cosine.
lim
x→ π2 −
6.
4
cos x
sin x
cos x
= lim
−
x→ π
2
4
=4
sin x
lim sec x − tan x = lim
x→ π2 −
7. lim
x→0+
x→ π2 −
1 − sin x LH
− cos x
= lim
=0
−
π
cos x
x→ 2 − sin x
1
x − ex + 1 LH
1 − ex
−ex
−1
1
LH
=
lim
=
lim
=
−
=
lim
ex − 1 x x→0+ xex − x
2
x→0+ ex + xex − 1
x→0+ 2ex + xex
8. lim+ x ln(x) = lim+
x→0
x→0
ln(x)
1
x
LH
= lim+
x→0
1
x
−1
x2
= lim+ −x = 0
x→0
1
= −∞
x
x→0
This limit is not an indeterminate since as x → 0+ the functions ln x → −∞ and
9. lim+ ln(x) −
1
x
→ ∞.
2
151 WebCalc Fall 2002-copyright Joe Kahlig
1
+ ln(x)
x→0 x
This limit is indeterminate since ∞ + −∞ = ∞ − ∞.
1
1 + x ln(x)
lim
+ ln(x) = lim
=∞
+
+
x
x→0 x
x→0
From question 8 we know that x ln x → 0 as x → 0+ . Thus we have the numerator going to 1
and the denominator going to 0.
10. lim+
11. lim
1+
x→∞
3
x
2x
3 2x
= lim eln(1+ x )
x→∞
Now compute lim 2x ln 1 +
x→∞
lim
x→∞
−6
x2
1+
3
x
Thus lim
x→∞
12. lim
x→∞
∗
6
−x2
= lim
x→∞
1
1+
1+
e3x + 2x
3
x
1
x
2x
3
x
3
x
lim 2x ln(1+ x3 )
= ex→∞
2 ln 1 +
= lim
3 2x
x→∞
1
3x +2x x
x→∞
2∗
LH
= lim
x→∞
−3
x2
1+ x3
−1
x2
=
=6
= lim eln(1+ x )
= lim eln(e
1
x
x→∞
3
x
lim 2x ln(1+ x3 )
= ex→∞
1
lim x ln(e
) = ex→∞
3x +2x
= e6
)
3e3x +2
1 ln e3x + 2x LH
3e3x + 2 LH
3x
Now compute lim ln e3x + 2x = lim
= lim e +2x = lim 3x
=
x→∞ x
x→∞
x→∞
x→∞ e
x
1
+ 2x
9e3x LH
27e3x
lim
=
lim
=3
x→∞ 3e3x + 2
x→∞ 9e3x
1
1
lim 1 ln(e3x +2x)
3x
x
x
Thus lim e3x + 2x = lim eln(e +2x) = ex→∞ x
= e3
x→∞
x→∞