If lim x tends to 1 [1-x^-(1/3)]/[1-x^-(2/3)] , then find the value ?

Step-by-step explanation:

lim x tends to 1 [1-x^-(1/3)]/[1-x^-(2/3)]

Find the value ?

Given that

lim x tends to 1 [1-x^-(1/3)]/[1-x^-(2/3)]

On taking [1-x^-(1/3)]/[1-x^-(2/3)]

We know that a^-n = 1/a^n

=> [1-(1/x⅓)]/[1-(1/x⅔)]

=> [(x⅓-1)/x⅓]/[(x⅔-1)/x⅔]

=> [(x⅓-1)/x⅓] × [x⅔/(x⅔-1)]

=> [(x⅓-1)x⅔] × [x⅓/(x⅔-1)]

=> [(x⅓-1)x⅓] / [(x⅔-1)]

Since a^m / a^n = a^(m-n)

=> [(x⅓-1)x⅓] / [(x⅓)²-1)]

=> [(x⅓-1)x⅓] / [(x⅓)²-1²)]

We know that

(a+b)(a-b) = a²-b²

=> [(x⅓-1)x⅓] / [(x⅓+1)(x⅓-1)]

On cancelling (x⅓-1)

=> (x⅓ / [(x⅓)+1)]

Now,

lim x tends to 1 [1-x^-(1/3)]/[1-x^-(2/3)]

=> lim x tends to 1 [(x⅓ / (x⅓)+1)]

=> 1⅓/(1⅓+1)

=> 1/(1+1)

=> 1/2

The value of the given problem is 1/2

Hope this helps..

the answer is 21%.