Friday

April 18, 2014

April 18, 2014

Posted by **kev** on Friday, March 2, 2007 at 8:28pm.

1.

Limit X approaching A

(X^1/3-a^1/3)/ x-a

2.

LiMIT x approaching 0

(1/3+x – 1/3) /x

On the first, would it help to write the denominator (x-a) as the difference of two cubes ((x^1/3 cubed - a^1/3 cubed)

second. use LHopitals rule. Take the derivative of the numerator, derivative of the denominator.

what if u never learned the lhopitals rule

All you have to do is to take the derivative of both numerator and denominator. That's L'Hopital's rule

If you never learned L'Hopital's rule, you should derive it first. Suppose that f(a) = g(a) = 0

Lim x--> a f(x)/g(x) =

Lim h--->0 f(a+h)/g(a+h) =

Lim h--->0

(f(a+h)- f(a))/(g(a+h) - g(a)) =

Lim h--->0

[(f(a+h)- f(a))/h] /(g(a+h) - g(a))/h] =

The limit of a quotient is the quotient of the limits provided both limits exists and the quotient of the limits also exists.

- limit calc question -
**Reiny**, Sunday, October 13, 2013 at 9:32pm1. consider the question to look like this

lim (x^(1/3) - a^(1/3) ) / ( (x^1/3)^3 - (a^1/3)^3) as x ----> a

=lim (x^1/3 - a^1/3)/[ (x^1/3) - a^1/3)(x^(2/3) +x^1/3 a^1/3 + a^2/3) ] as x--->a

= lim 1/ (x^1/3) - a^1/3)(x^(2/3) +x^1/3 a^1/3 + a^2/3) as x --> a

= 1/(a^2/3 + a^1/3 a^1/3 + a^2/3)

= 1/(3 a^(2/3)

2. without L'Hopital's rule

lim (1/(3+x) - 1/3 )/x , as x-->0

= lim [(3 - 3 - x)/(3(3+x)) / x

= lim [ -x/(3(3+x))/x

= lim -1/(3(3+x)) as x-->0

= -1/9

- limit calc question -
**Reiny**, Sunday, October 13, 2013 at 9:32pm1. consider the question to look like this

lim (x^(1/3) - a^(1/3) ) / ( (x^1/3)^3 - (a^1/3)^3) as x ----> a

=lim (x^1/3 - a^1/3)/[ (x^1/3) - a^1/3)(x^(2/3) +x^1/3 a^1/3 + a^2/3) ] as x--->a

= lim 1/ (x^1/3) - a^1/3)(x^(2/3) +x^1/3 a^1/3 + a^2/3) as x --> a

= 1/(a^2/3 + a^1/3 a^1/3 + a^2/3)

= 1/(3 a^(2/3)

2. without L'Hopital's rule

lim (1/(3+x) - 1/3 )/x , as x-->0

= lim [(3 - 3 - x)/(3(3+x)) / x

= lim [ -x/(3(3+x))/x

= lim -1/(3(3+x)) as x-->0

= -1/9

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