Unformatted text preview: δx → 0
δx → 0 δx → 0 Limit from the RIGHT
L imit { f(x) } = L imit
x → 0+ x → 0+ { 1 x } = +1 0 = +∞ , something undefined.
/
/ Limit { f(x) } = Limit {f(0 + δ x)} = Limit { 1 (0 + δ x) } =+1 0 = +∞
/
/
δx → 0
δx → 0
δx → 0
The LIMIT from the left does not exist. And, the LIMIT from the right does not exist.
left
right
Also Value { f(x) } = f(0) = 1 0 = ∞ , something undefined.
/
x=0
So on any one of the three counts, be it LIMIT from the left or LIMIT from the right
left
right
or the VALUE, f(x) is discontinuous at x = 0. Moreover, we cannot remove this
discontinuous
discontin
discontinuity. 50 sin(x)
continuous at x = 0 ?
continuous
x
We shall use the trigonometric identity sin (A ± B) = sin A cos B ± cos A sin B.
Also, when x is very small, we may say : sin (x) = x . Example 7: Is the function f(x) = Limit from the LEFT
sin(− δ x)
L imit {f(x)} = L imit { sin(0 − δ x) } = L imit {
}
δx → 0
δx → 0
δx → 0
− δx
(0 − δ x)
− δx
L
L
} = δ ximit0 {1} = 1 .
= δ ximit0 {
→
→
− δx irst
take
Note how we simplify ffirst and then take the limit. We can simplify first because
simplify ir
δx only TENDS TO zero. δx ≠ 0.
Limit from the RIGHT
L imit {f(x)} = L imit { sin(0 + δ x) } = L imit { sin(+ δ x) }
δx → 0
δx → 0
δx → 0
+ δx
(0 + δ x)
δx
} = L imit {1} = 1 .
δx → 0
δx → 0
δx
Note how we simplify ffirst and then take the limit. We can simplify first because
irst
take
simplify ir
δx only TENDS TO zero. δx ≠ 0.
= L imit { Since, L imit  { f(x) } = 1 = L imit+ { f(x) } , we say L imit
x→0 x→0 x→0 sin(x)
{
} =1.
x Value { f(x) } = Value { sin(x) } = 0 , which is something undefined.
/0
x
x=0
x=0
Since L imit f(x) ≠ Value f(x) we say f(x) is discontinuous at x = 0.
discontinuous
x→0
x=0
sin(x) for x ≠ 0
We may remove this discontinuity by redefining f(x) to be : f(x) ={ x
1 for x = 0
51 Example 8 : Is the function f(x) defined below continuous anywhere ?
continuous
f(x) = +1 if x is rational
rational { 0 if x is irrational
irr
ir We saw t...
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 Fall '09
 TAMERDOğAN
 Limit, Δx

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