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Calculus

Calculus 1000A/B

Chris Brandl

Fall

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MATH 127 Fall 2012
Assignment 5 Solutions
1. Evaluate each of the following limits
11x
(a) lim p
x!▯1 x + 4x
11x 11x
lim p = lim q ▯ ▯
x!▯1 x + 4x x!▯1 x2 1 +4x
x2
11x
= lim q
x!▯1 p 2 4
x 1 + x
11x
= x!▯1 q
jxj 1 + x
11x
= lim q
x!▯1 ▯x 1 + 4
x
▯11
= lim q
x!▯1 4
1 +x
r !
4
= ▯11 note that lim 1 + = 1
x!▯1 x
3e + e▯x
(b) lim
x!▯1 e ▯ e▯x+1
3e + e▯x 3e + x
lim = lim e
x!▯1 e ▯ e▯x+1 x!▯1 e ▯ ex
2x
3ex+1
= lim 2x
x!▯1 ex▯e
e
3e2x+ 1
= lim 2x
x!▯1 ▯ ▯ e ▯
1
= ▯ since lim 3e2x= 0 and lim e2x= 0
e x!▯1 x!▯1
▯x p
(c) lim e sin x + 2
x!1 p p
Note that ▯1 ▯ sin x + 2 ▯ 1 so we get that ▯ex ▯ e▯xsin x + 2 ▯ e▯x.
▯x ▯x p
Since lim ▯e = 0 we get by the Squeeze Theorem that lim sin x + 2 = 0.
x!1 x!1
1 p 2
(d) lim 4x ▯ 2x + 1 ▯ 2x
x!1
▯ p ▯
p ▯p ▯ 4x ▯ 2x + 1 + 2x
lim 4x ▯ 2x + 1 ▯ 2x = lim 4x ▯ 2x + 1 ▯ 2x p
x!1 x!1 4x ▯ 2x + 1 + 2x
2 2
= lim px ▯ 2x + 1 ▯ 4x
x!1 4x ▯ 2x + 1 + 2x
▯2x + 1
= x!1 p 2
4x ▯ 2x + 1 + 2x
▯2x + 1
= lim q ▯ ▯
x!1 x2 4 ▯ 2+ 1 + 2x
x x2
▯2x + 1
= lim q
x!1 2 1
jxj 4 ▯ x+ x2+ 2x
= lim q ▯2x + 1
x!1 2 1
x 4 ▯x + x2+ 2x
▯ ▯
x ▯2 + x
= lim ▯q ▯
x!1 x 4 ▯ 2+ 1 + 2
x x
1
▯2 + x
= x!1 q 2 1
4 ▯x + x2+ 2
▯2
= p
4 + 2
1
= ▯
2
2. For the following functions, using the de▯nition of the derivative either ▯nd the indi-
cated derivative or show the derivative does not exist.
8
>
< x x < 0
(a) f (0) where f(x) =
:
sin(x) x ▯ 0
0 f(x) ▯ f(0)
We need to compute f (0) = lim but since f is de▯ned di▯erently on
x!0 x ▯ 0
2 either side of x = 0 we must observe left and right hand limits.
f(x) ▯ f(0) x ▯ sin(0)
li▯ = lim▯
x!0 x x!0 x
x
= x!0▯x
= 1
f(x) ▯ f(0) sin(x) ▯ sin(0)
li+ = lim+
x!0 x x!0 x
sin(x)
= lim+
x!0 x
= 1
f(x) ▯ f(0) f(x) ▯ f(0)
Since lim = 1 = lim we get that
x!0+ x x!0▯ x
0 f(x) ▯ f(0)
f (0) = lim = 1
x!0 x
0 1
(b) f (x) where f(x) 1 + x
f (x) = limx + h) ▯ f(x)
h!0 h
1 1
p1+x+h▯ p1+x
= lim
h!0p h p
1 + x ▯ 1 + x + h
= lim p p
h!0h( 1 + x + h 1 + x)
▯ p p ▯▯ p p ▯
1 + x ▯ 1 + x + h 1 + x + 1 + x + h
= h!0 p p p p
h( 1 + x + h 1 + x) 1 + x + 1 + x + h
1 + x ▯ (1 + x + h)
= lim p p
h!0h 1 + x + h(1 + x) + h(1 + x + h) 1 + x
= lim p ▯h p
h!0h 1 + x + h(1 + x) + h(1 + x + h) 1 + x
▯1
= h!0p p
1 + x + h(1 + x) + (1 + x + h) 1 + x
▯1
= p p
1 + x(1 + x) + (1 + x) 1 + x
▯1
= 3
2(1 + x)
3 (c) f (x) where f(x) = x ▯ 3x
0 f(x + h) ▯ f(x)
f (x) = lim
h!0 h

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