Transcript Ch.2 Limits and derivatives

Example
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x3
.
Ex. Find all asymptotes of the curve y  2
x  2x  3
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Sol. x 2  2 x  3  ( x  3)( x  1)  lim y  lim y  
x 3
x 1
So x=3 and x=-1 are vertical asymptotes.
m  lim y / x  1, b  lim( y  mx)  lim( y  x)  2
x 
x 
So y=x+2 is a slant asymptote.
x 
Example
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| x|
Ex. Find asymptotes of the curve f ( x) 
.
x 1
Sol. vertical asymptote x  1
horizontal asymptote y  1, y  1
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( x  2)2
.
Ex. Find asymptotes of the curve y 
2( x 1)
Sol. vertical asymptote x  1
x 3
slant asymptote y  
2 2
Curve sketching
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A. Domain
B. Intercepts
C. Symmetry
D. Asymptotes
E. Intervals of increase or decrease
F. Local maximum and minimum values
G. Convexity and points of inflection
H. Sketch the curve
Example
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 x2
Ex. Sketch the graph of y  e .
Sol. A. The domain is (-1,+1). B. The y-intercept is 1.
C. f is even. D. asymptotes: y=0 is horizontal asymptote.
 x2
E. f ( x)  2xe , when x>0, f ( x)  0, so f(x) decreasing
in (0,+1) and increasing in (-1,0).
F. x=0 is local and global maximum point.
2 2
 x2
2

G. f ( x)  2e (2x 1), f(x) concave in ( , ) and
2 2
convex otherwise
Example
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2 x2
Sketch the graph of y 
.
2
(1  x)
Example
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Ex. Prove the inequality: ( x2 1)ln x  ( x 1)2 ( x  0).
Proof. Let f ( x)  ( x2 1)ln x  ( x 1)2 , then
1

f ( x)  2 x ln x  x  2 
x
1
f ( x)  2ln x  1  2
x
2
2( x  1)
f ( x) 
x3
Indeterminate forms
ln x
Question: find the limit lim
we can’t apply the limit
x 1 x  1
law because the limit of the denominator is 0. In fact the limit
of the numerator is also 0. We call this type of limit an
indeterminate form.
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Generally, if both f ( x)  0 and g ( x)  0 as x  a ,
then the limit
f ( x)
lim
xa g ( x)
may or may not exist and is called an indeterminate form of
type 0/0
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Previous methods
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For rational functions, for example,
x2  x
x( x  1)
x
1
lim 2
 lim
 lim
 .
x 1 x  1
x 1 ( x  1)( x  1)
x 1 x  1
2
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sin x
1
x 0
x
The important limit: lim
Does not work for general cases. There is a systematic
method, known as L’Hospital’s Rule, for evaluation of
indeterminate forms.
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L’Hospital’s rule
L’Hospital’s Rule Suppose f and g are differentiable and
g ( x)  0 near a (except possibly at a). Assume that
lim g ( x )  0
and
lim f ( x )  0
xa
xa
or that
lim f ( x)   and lim g ( x)  
Then
xa
xa
f ( x)
f ( x)
lim
 lim
x a g ( x)
x a g ( x )
if the last limit exists (can be a real number or  or  ).
Remarks
Remark1. L’Hospital’s Rule can be used to evaluate the
indefinite limit of type 0/0 or 1/1.
Remark2. L’Hospital’s Rule is also valid when “x!a” is
replaced by x!a+, x!a-, x!+1, x!-1.
f ( x)
Remark3. If lim
is still an indeterminate type, we can
x  a g ( x)
use L’Hospital’s Rule again.
Examples
x  sin x
Ex. Find lim
.
3
x 0
x
x  sin x
1  cos x
sin x 1
 lim
 lim
 .
Sol. lim
3
2
x 0
x 0
x 0 6 x
x
3x
6
xn
Ex. Find lim x (a  1).
x  a
xn
nxn1
n!
Sol. lim
 lim x
  lim x
 0.
x
n
x  a
x  a ln a
x  a (ln a)
Note: This example indicates that exponential infinity is
much bigger than any power infinity.
When not to use L’Hospital’s rule
1
x sin
x.
lim
Ex. Find x0
sin x
2
Sol.
1
1
1
x sin
2 x sin  cos
x  lim
x
x
lim
x 0 sin x
x 0
cos x
2
L’Hospital’s Rule gives nothing! Correct solution is 0.
Other indeterminate types
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There are some other indeterminate types which can be
changed into 0/0 or 1/1 type: 0¢1, 1-1, 11, 10, 00.
Ex. Find lim x ln x.
x 0
1
ln x
Sol. lim x ln x  lim
 lim x  0.
x 0 
x 0  1
x 0 
1
 2
x
x
Homework 10
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Section 4.4: 22, 30, 31, 45, 48, 55, 74
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Section 4.5: 26, 28, 64