#lim_(trarr1){t-1}/(t^3-1)=# ?

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Answer 1 · 2016-07-21T17:13:28

#1/3#

Making #x=z^3#

#lim_(xrarr1) (root3x-1)/(x-1) equiv lim_{z->1}(z-1)/(z^3-1)#

but

#(z-1)/(z^3-1)=1/(z^2+z+1)#

so

#lim_(xrarr1) (root3x-1)/(x-1) =1/3#

Answer 2 · 2016-07-21T17:22:55

#1/3#.

In fact, it is a Standard Form of Limit that

#lim_(xrarra)(x^n-a^n)/(x-a)=n*a^(n-1), where, n in RR#.

In our example, #a=1, n=1/3#.

The Reqd. Limit#=1/3*(1)^(1/3-1)=1/3#.

Aliter :-

Take substn. #x=t^3#, so that, as #xrarr1, trarr1#.

#:.# Reqd. Limit#=lim_(trarr1){root(3)(t^3)-1}/(t^3-1)#

#=lim_(trarr1)(t-1)/{(t-1)(t^2+t+1)}#

As #trarr1, t!=1#, so, #(t-1)!=0# can be cancelled, to get,

The Reqd. Lim.#=lim_(trarr1)1/(t^2+t+1)=1/3#, as before!

Enjoy Maths.!