Evaluate the limits

Question

Evaluate lim ((x^(1/n)-a^(1/n))/(x-a)) as x->a and lim ((x^n-a^n)/(x-a)) as x->a?

 

I didnt learn l'hopital's rule

 

Answer 1

The first limit is the derivative of the function  x^1/n  evaluated at x = a. 

The derivative of x^1/n  is   (1/n) x^{1/n -1}    so the limit is    (1/n) a^{1/n -1}  = (1/n) / a^1-1/n.

 

The second limit is the derivative of the function xⁿ  evaluated at x =a.

The derivative of xⁿ is n n x^n-1   so the limit is   n a^n-1 .

 

In the case of the second limit, it is easy to see how this works by noting that

xⁿ - aⁿ  =   (x -a) (  x^n-1 a⁰ + x^n-2 a¹  +   ...  +  x⁰ a^n-1)   

All of the terms inside the second set of parentheses evaluate ( in the limit x ->n) to a^n-1

There are n of them, so in the limit  xⁿ - aⁿ =  (x -a) n a^n-1

Thus the desired limit is n a^n-1.

 

A similar, but more complicated, analysis leads the the result above for the first limit.

 

 

 

Answer 2

To make the math easier, substitute y for x^1/n and b for a^1/n

 

So (x^1/n - a^1/n)/(x-a) becomes:

(y - b) / (yⁿ - bⁿ)

 

Factor the denominator:

(y - a) / (y - a) (y^n-1 + by^n-2 + b²y^n-3 + ... b^n-2y + b^n-1))

 

Divide through by (y-a)

1 / (y^n-1 + by^n-2 + b²y^n-3 + ... b^n-2y + b^n-1))

 

As x-->a, x^1/n --> a^1/n, so y --> b

 

So the above expression approaches:

1 / (b^n-1 + bb^n-2 + b²b^n-3 + ... b^n-2b + b^n-1))

 

All the terms in the denominator equal b^n-1, and there are n of them.  So that expression approaches:

1 / (nb^n-1)

 

And that is the limit.