What is the limit of (t^(1/3)-2)/(t-8)t132t8 as t approaches 8?

2 Answers
Sep 13, 2016

" The Limit="1/12 The Limit=112.

Explanation:

There is a Standard Form of Limit :-

lim_(xrarra) (x^n-a^n)/(x-a)=n*a^(n-1).

Using this Form, we have,

"The Limit="lim_(trarr8) (t^(1/3)-2)/(t-8)

=lim_(trarr8) (t^(1/3)-8^(1/3))/(t-8)

=1/3*8^(1/3-1)

=1/3*(2^3)^(-2/3)=1/3*2^-2=1/3*1/2^2

:." The Limit="1/12, tallying with Claude D. 's Answer!

Sep 13, 2016

1/12.

Explanation:

We subst. t=(x+2)^3," with a note that, as "trarr8, xrarr0.

Hence, the Reqd. Limit=lim_(xrarr0) {((x+2)^3)^(1/3)-2}/{(x+2)^3-8}.

Here, we use, (a+b)^3=a^3+b^3+3ab(a+b), to get,.

The Limit=lim_(xrarr0) {(x+2)-2}/{x^3+8+3*x*2(x+2)-8}

=lim_(xrarr0) x/(x^3+6x^2+12x)

=lim_(xrarr0) cancelx/{cancelx(x^2+6x+12)}

=lim_(xrarr0) 1/(x^2+6x+12)

=1/12, as before we had!

Enjoy Maths.!