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

Question

4868 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-13T15:56:47

#" The Limit="1/12#.

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!

Answer 2 · 2016-09-13T16:15:13

#1/12#.

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.!