07 - Limit of a Function
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Transcript of 07 - Limit of a Function
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8/11/2019 07 - Limit of a Function
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Consider the graph of
f(x) =x2
+ 1
Observe the values of
f(x) asxapproaches 3
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Observe the values off(x) =x2+ 1asxapproaches 3:
We say that
limx3
f(x)10
x 2 2.5 2.9 2.99 2.999 3 3.001 3.01 3.1 3.5 4
f (x) 5 7.25 9.41 9.94 9.994 10 10.006 10.06 10.61 13.25 17
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The concept of a l m tis a means by which todescribe the behavior of a function as the
independent variable xgets very close to afixed number, say .
Limit off(x)asxapproaches athe value that
f(x)approaches as
xgets closer to
a In symbols:
)(lim xfax
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Rule 1 The limit of a constant function at anynumber is equal to its constant value.
Rule 2 Iff is a linear function, then
Rule 3 The limit of a constant multiple of afunction is equal to the product of theconstant and the limit of the function
limx a kk
bmabmxxfaxax
)(lim)(lim
limxa
kg(x)klimxa
g(x)
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Rule 4 The limit of a sum is the sum of limits.
Rule 5 The limit of a product is the product ofthe limits.
Rule 6 The limit of the nth power of a functionis the nth power of the limit of the function.
limxa
f(x)g(x) limxa
f(x) limxa
g(x)
)(lim)(lim)()(lim xgxfxgxfaxaxax
nax
n
axxfxf )(lim)(lim
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Rule 7 The limit of a sum is the sum of limits.
Rule 8 The limit of the nthroot of a function isthe nthroot of the limit of the function. Thelimit of a product is the product of the limits.
limxa
f(x)g(x) limxa
f(x) limxa
g(x)
nax
n
axxfxf )(lim)(lim
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0
0
1. To evaluate the limit of the function,
just substitute the value a to x and
then evaluate the function.
2. If upon substitution of a to x in thefunction, the value becomes it is an
indication that the function has to be
simplified first either by factoring or
by rationalizing the numerator ordenominator, and then canceling the
common factor or factors.
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Find the limit, if it exists.
1.
2.
3.
4.
5.62
32
3
lim
x
xx
x
)1(lim 22
xxx
)75(lim 231
xxxx
limx4
x2 162x 8
43
12lim
2
34
2
x
xx
x
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Find the limit, if it exists.
6. 11.
7.
8.
9.
10.4
82
3
2
lim
x
x
x
3
273
3lim
x
x
x
1
1
lim1
x
x
x
7
492
7lim
x
x
x
limx2
x2 4
x2 x2
1
13
1lim
x
x
x
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Given . Find the limit
of f(x) asxapproaches 2..
limx2 f(x) 5
x 1 1.5 1.9 1.99 1.999 2 2.001 2.01 2.1 2.5 3
f (x) 3 4 4.8 4.98 4.998 4 4.004 4.04 4.41 6.25 9
f(x)2x1 for x 2
x 2 for x 2
limx2 f(x) 4
Therefore, lim
x2f(x)does not exist.
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-1
1
-2-3 -1
-3
3
Limit does not exist.
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1 2 3
1
5
)(lim1
xfx
)(lim1
xfx
)(
lim1xf
x
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Evaluate the limit, if it exists.
a.
b.
33
33,9
3,5
)( 2
xifx
xifx
xifx
xf
)(lim3
xfx
)(
lim3xf
x
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Evaluate the limit, if it exists.
1.
1,2
1,4)(2
2
xifx
xifxxh
)(lim1 xhx
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Evaluate the limit, if it exists.
xx
xlim
0
x
x
xlim0
x
x
x
lim0
6
4
2
-2
-4
-5 5
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Evaluate the limit, if it exists. 1.
x
x
x 1
22lim
1
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