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Published byAlexandra Welch Modified over 9 years ago
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2.5 β Continuity A continuous function is one that can be plotted without the plot being broken. Is the graph of f(x) a continuous function on the interval [0, 4]? No At what values of x is the function discontinuous and why? π₯=1 πβπππ ππ π ππ’ππ. π₯=2 πβπππ ππ π βπππ. π₯=4 πβπππ ππ π ππ’ππ. Is the graph of f(x) continuous at π₯=3? Yes
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What are the rules for continuity at a point?
lim π₯β 3 β π π₯ = lim π₯β π π₯ = 2 2 lim π₯β3 π π₯ = π 3 = 2 2 lim π₯β 1 β π π₯ = lim π₯β π π₯ = 1 lim π₯β1 π π₯ = π 1 = π·ππΈ 1 lim π₯β 2 β π π₯ = 1 lim π₯β π π₯ = 1 lim π₯β2 π π₯ = 1 π 2 = 2 What are the rules for continuity at a point? lim π₯β 4 β π π₯ = 1 lim π₯β π π₯ = ππππ lim π₯β4 π π₯ = π 4 = ππππ 0.5
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2.5 β Continuity
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2.5 β Continuity ο ο π₯=1 π 1 =1 π π ππ₯ππ π‘π lim π₯βπ π π₯ ππ₯ππ π‘π
π π₯ ππ πππ‘ ππππ‘πππ’ππ’π ππ‘ π₯=1. β΄
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2.5 β Continuity ο ο ο π₯=2 π π ππ₯ππ π‘π π 2 =2 lim π₯βπ π π₯ ππ₯ππ π‘π
2β 1 π π₯ ππ πππ‘ ππππ‘πππ’ππ’π ππ‘ π₯=2. β΄
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2.5 β Continuity ο ο ο π₯=3 π π ππ₯ππ π‘π π 2 =2 lim π₯βπ π π₯ ππ₯ππ π‘π
2=2 π π₯ ππ ππππ‘πππ’ππ’π ππ‘ π₯=3. β΄
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2.5 β Continuity ο ο ο π₯=0 (left end point) π π ππ₯ππ π‘π π 0 =1
lim π₯β π + π π₯ ππ₯ππ π‘π lim π₯β π π₯ =1 ο lim π₯β π + π π₯ =π(π) 1=1 β΄ π π₯ ππ ππππ‘πππ’ππ’π ππ‘ π‘βπ ππππ‘ ππππππππ‘, π₯=0.
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2.5 β Continuity ο ο ο π₯=4 (right end point) π π ππ₯ππ π‘π π 4 =0.5
lim π₯β π β π π₯ ππ₯ππ π‘π lim π₯β 4 β π π₯ =1 ο lim π₯β π + π π₯ =π(π) 0.5β 1 β΄ π π₯ ππ πππ‘ ππππ‘πππ’ππ’π ππ‘ π‘βπ πππβπ‘ ππππππππ‘, π₯=4.
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Removable Discontinuity
Removable discontinuity occurs at a point where the function has a hole but does not have a function value. πΆππππ‘π π π ππππππ ππ’πππ‘πππ, π π₯ , π‘βππ‘ ππ ππππ‘πππ’ππ’π ππ‘ π₯=2. lim π₯β2 π π₯ ππ₯ππ π‘π lim π₯β2 π π₯ =1 lim π₯β2 π π₯ =π(2) 1=1 ο· π 2 ππ₯ππ π‘π π 2 =1 π(π₯) ππ π₯β 2 1 ππ π₯=2 π π₯ = π π₯ ππ πππ ππππ‘πππ’ππ’π ππ‘ π₯=2 π΄ βπππ ππ₯ππ π‘ ππ‘ π₯=2. π π₯ ππ ππππ‘πππ’ππ’π ππ‘ π₯=2. π
ππππ£ππππ πππ ππππ‘πππ’ππ‘π¦ ππ‘ π₯=2.
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Removable Discontinuity
Removable discontinuity occurs at a point where the function has a hole but does not have a function value. πΆππππ‘π π π ππππππ ππ’πππ‘πππ, π π₯ , π‘βππ‘ ππ ππππ‘πππ’ππ’π ππ‘ π₯=3. π(π₯) lim π₯β3 π π₯ ππ₯ππ π‘π lim π₯β3 π π₯ =0 ο· lim π₯β3 π π₯ =π(3) 0=0 π 3 ππ₯ππ π‘π π 3 =0 π π₯ ππ πππ ππππ‘πππ’ππ’π ππ‘ π₯=3 π(π₯) ππ π₯β 3 0 ππ π₯=3 π π₯ = π΄ βπππ ππ₯ππ π‘ ππ‘ π₯=3. π π₯ ππ ππππ‘πππ’ππ’π ππ‘ π₯=3. π
ππππ£ππππ πππ ππππ‘πππ’ππ‘π¦ ππ‘ π₯=3.
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Removable Discontinuity
Example: The given function is discontinuous. Where is it discontinuous and is it removable? π π₯ = π₯ 2 β4 π₯β2 π₯β2=0 π₯=2 π(π₯) ππ πππ ππππ‘πππ’ππ’π ππ‘ π₯=2 π π₯ = π₯β2 π₯+2 π₯β2 πβπ ππππ‘ππ π₯β2 πππππππ . πβπππ ππ π βπππ ππ‘ π₯=2. πΌπ‘ ππ πππππ£ππππ ππ‘ π₯=2.
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Removable Discontinuity
Example: The given function is discontinuous. Where is it discontinuous and is it removable? π π₯ =π‘ππ π 2 π₯ π 2 π₯= π 2 , 3π 2 , 5π 2 , β― π₯=1, 3, 5, 7, β― π π₯ = π ππ π 2 π₯ πππ π 2 π₯ π π₯ ππ πππ ππππ‘πππ’ππ’π ππ‘ π₯=1,3, 5, 7, β― π π₯ = π ππ π 2 π₯ πππ π 2 π₯ πππ π 2 π₯ =0 πβπ ππππ‘ππ πππ π 2 π₯ ππππ πππ‘ ππππππ. πππ‘ π= π 2 π₯ πππ π=0 πβπππ ππ ππ πππππ£ππππ πππ ππππ‘πππ’ππ‘π¦ πππ π(π₯) π= π 2 , 3π 2 , 5π 2 , β―
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Removable Discontinuity
Examples
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2.5 β Continuity π π₯ =2 π₯ 3 β16 π₯ 2 +38π₯β22 1,5
π π₯ =2 π₯ 3 β16 π₯ 2 +38π₯β ,5 π π₯ ππ π ππππ¦πππππππ β΄ππππ‘πππ’ππ’π π 1 = 2 π 5 = 18 π π₯ =8 8=2 π₯ 3 β16 π₯ 2 +38π₯β22 π₯=4.547
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2.6 β Limits Involving Infinity; Asymptotes of Graphs
π΄π π₯β 0 + , π¦ββ lim π₯β π π₯ =β π΄π π₯βββ, π¦β0 π΄π π₯ββ, π¦β0 lim π₯βββ π π₯ =0 lim π₯ββ π π₯ =0 π΄π π₯β 0 β , π¦βββ lim π₯β 0 β π π₯ =ββ
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2.6 β Limits Involving Infinity; Asymptotes of Graphs
π».π΄. ππ‘ π¦=5 π».π΄. ππ‘ π¦=0 π.π΄. ππ‘ π₯=4 lim π₯ββ π π₯ =β2 lim π₯β π π₯ =β lim π₯β β7 + π π₯ =β π».π΄. ππ‘ π¦=β2 π.π΄. ππ‘ π₯=2 π.π΄. ππ‘ π₯=β7
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2.6 β Limits Involving Infinity; Asymptotes of Graphs
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2.6 β Limits Involving Infinity; Asymptotes of Graphs
Examples
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2.6 β Limits Involving Infinity; Asymptotes of Graphs
Oblique Asymptotes
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