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

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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.!

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