Try it yourself! 3 x + 3 √ x − 1 Compute lim . 2 x − 2 x → 2
Try it yourself! 3 x + 3 √ x − 1 Compute lim . 2 x − 2 x → 2 After checking that 2 is in the domain of the expression, we just plug it in: 6 + 3 √ 2 − 1 = 7 lim 4 − 2 2 x → 2
Continuity x 2 − 5 x Example. Compute lim x − 5 . x → 5
Continuity x 2 − 5 x Example. Compute lim x − 5 . x → 5 After checking that 5 is in the domain of the expression,
Continuity x 2 − 5 x Example. Compute lim x − 5 . x → 5 After checking that 5 is in the domain of the expression, We discover that it’s not!
Continuity x 2 − 5 x Example. Compute lim x − 5 . x → 5 After checking that 5 is in the domain of the expression, We discover that it’s not! So we have to do something other than just plugging it in.
Independence of the value
Independence of the value If f ( x ) = g ( x ) except possibly at x = a
Independence of the value If f ( x ) = g ( x ) except possibly at x = a then x → a f ( x ) = lim lim x → a g ( x )
Independence of the value If f ( x ) = g ( x ) except possibly at x = a then x → a f ( x ) = lim lim x → a g ( x ) More precisely, if one limit is defined, so is the other, and in this case they are equal.
Independence of the value If f ( x ) = g ( x ) except possibly at x = a then x → a f ( x ) = lim lim x → a g ( x ) More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a .
Independence of the value If f ( x ) = g ( x ) except possibly at x = a then x → a f ( x ) = lim lim x → a g ( x ) More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a . Note: f nor g need not have the same domain, and need not be defined at a .
Independence of the value If f ( x ) = g ( x ) except possibly at x = a then x → a f ( x ) = lim lim x → a g ( x ) More precisely, if one limit is defined, so is the other, and in this case they are equal. This is true because the limit only takes into account values near but not at a . Note: f nor g need not have the same domain, and need not be defined at a . One needs only f ( x ) = g ( x ) over some [ b , a ) ∪ ( a , c ].
Independence of the value x 2 − 5 x Example. Compute lim x − 5 . x → 5
Independence of the value x 2 − 5 x Example. Compute lim x − 5 . x → 5 x 2 − 5 x lim x − 5 x → 5
Independence of the value x 2 − 5 x Example. Compute lim x − 5 . x → 5 x 2 − 5 x x ( x − 5) lim = lim x − 5 x − 5 x → 5 x → 5
Independence of the value x 2 − 5 x Example. Compute lim x − 5 . x → 5 x 2 − 5 x x ( x − 5) lim = lim = lim x → 5 x x − 5 x − 5 x → 5 x → 5
Independence of the value x 2 − 5 x Example. Compute lim x − 5 . x → 5 x 2 − 5 x x ( x − 5) lim = lim = lim x → 5 x = 5 x − 5 x − 5 x → 5 x → 5
Independence of the value x 2 − 5 x Example. Compute lim x − 5 . x → 5 x 2 − 5 x x ( x − 5) lim = lim = lim x → 5 x = 5 x − 5 x − 5 x → 5 x → 5 Here in the second equality, we used that x ( x − 5) = x for any value x − 5 of x other than 5.
Independence of the value x − 2 Example. Compute lim √ . x 2 − 3 − 1 x → 2
Independence of the value x − 2 Example. Compute lim √ . x 2 − 3 − 1 x → 2 Note we cannot just plug in x = 2,
Independence of the value x − 2 Example. Compute lim √ . x 2 − 3 − 1 x → 2 Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist,
Independence of the value x − 2 Example. Compute lim √ . x 2 − 3 − 1 x → 2 Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist, the limit of the denominator is 0.
Independence of the value x − 2 Example. Compute lim √ . x 2 − 3 − 1 x → 2 Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist, the limit of the denominator is 0. Remark: If the limit of the numerator were not zero,
Independence of the value x − 2 Example. Compute lim √ . x 2 − 3 − 1 x → 2 Note we cannot just plug in x = 2, because even though the limit of the numerator and denominator both exist, the limit of the denominator is 0. Remark: If the limit of the numerator were not zero, the limit of the quotient would be infinite or undefined. But it is zero.
We want to cancel some zeroes from the numerator and denominator.
We want to cancel some zeroes from the numerator and denominator. We start by removing the square-root from the denominator: √ x 2 − 3 + 1 x − 2 x − 2 lim √ = lim √ · √ x 2 − 3 − 1 x 2 − 3 − 1 x 2 − 3 + 1 x → 2 x → 2 √ x 2 − 3 + 1) ( x − 2)( = lim ( x 2 − 3) − 1 x → 2 √ x 2 − 3 + 1) ( x − 2)( = lim x 2 − 4 x → 2 √ x 2 − 3 + 1) ( x − 2)( = lim ( x − 2)( x + 2) x → 2
Now we are able to cancel:
Now we are able to cancel: √ √ x 2 − 3 + 1) x 2 − 3 + 1 ( x − 2)( lim = lim ( x − 2)( x + 2) x + 2 x → 2 x → 2
Now we are able to cancel: √ √ x 2 − 3 + 1) x 2 − 3 + 1 ( x − 2)( lim = lim ( x − 2)( x + 2) x + 2 x → 2 x → 2 Finally, we can plug in 2.
Now we are able to cancel: √ √ x 2 − 3 + 1) x 2 − 3 + 1 ( x − 2)( lim = lim ( x − 2)( x + 2) x + 2 x → 2 x → 2 Finally, we can plug in 2. √ √ x 2 − 3 + 1 2 2 − 3 + 1 = 1 lim = x + 2 2 + 2 2 x → 2
Try it yourself! Compute: √ x 2 − 5 − 2 lim x − 3 x → 3
Limits from one-sided limits
Limits from one-sided limits If lim x → x − 0 f ( x ) and lim x → x + 0 f ( x ) exist
Limits from one-sided limits If lim x → x − 0 f ( x ) and lim x → x + 0 f ( x ) exist and are equal
Limits from one-sided limits If lim x → x − 0 f ( x ) and lim x → x + 0 f ( x ) exist and are equal then also lim x → x 0 f ( x ) exists
Limits from one-sided limits If lim x → x − 0 f ( x ) and lim x → x + 0 f ( x ) exist and are equal then also lim x → x 0 f ( x ) exists and lim f ( x ) = lim x → x 0 f ( x ) = lim f ( x ) x → x + x → x − 0 0
Limits from one-sided limits If lim x → x − 0 f ( x ) and lim x → x + 0 f ( x ) exist and are equal then also lim x → x 0 f ( x ) exists and lim f ( x ) = lim x → x 0 f ( x ) = lim f ( x ) x → x + x → x − 0 0 All the limit laws hold for one-sided limits as well.
Limits from one-sided limits We find the one-sided limits lim x → 0+ f ( x ) = 0 and lim x → 0 − f ( x ) = 0 by direct substitution.
Limits from one-sided limits We find the one-sided limits lim x → 0+ f ( x ) = 0 and lim x → 0 − f ( x ) = 0 by direct substitution. Thus lim x → 0 f ( x ) = 0.
Try it yourself Find lim x → 1 f ( x ), where � x 2 − 1 x > 1 x − 1 f ( x ) = x 2 + 1 x < 1
Squeeze theorem
Squeeze theorem If f ( x ) ≤ g ( x ) ≤ h ( x ) on some ( a , b ) ∪ ( b , c )
Squeeze theorem If f ( x ) ≤ g ( x ) ≤ h ( x ) on some ( a , b ) ∪ ( b , c ) and x → b f ( x ) = lim lim x → b h ( x )
Squeeze theorem If f ( x ) ≤ g ( x ) ≤ h ( x ) on some ( a , b ) ∪ ( b , c ) and x → b f ( x ) = lim lim x → b h ( x ) Then lim x → b g ( x ) exists and x → b f ( x ) = lim lim x → b g ( x ) = lim x → b h ( x ) Also sometimes called the sandwich theorem
Squeeze theorem If f ( x ) ≤ g ( x ) ≤ h ( x ) on some ( a , b ) ∪ ( b , c ) and x → b f ( x ) = lim lim x → b h ( x ) Then lim x → b g ( x ) exists and x → b f ( x ) = lim lim x → b g ( x ) = lim x → b h ( x ) Also sometimes called the sandwich theorem or the two policemen and one drunk theorem.
If a police officer stands on each side of a drunk, and both police go to the jail, then no matter how the drunk wobbles about, he will also end up in jail.
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