Question

z31210. lim

Answer 1

Let the function be

f( z )= ( z 1 3 −1 ) ( z 1 6 −1 )

We have to find the value of the function at Limit z→1

So we need to check the function by substituting the value at particular point that it should not be of the form 0 0

f( z )= ( 1 1 3 −1 ) ( 1 1 6 −1 ) = ( 1−1 ) 1−1 = 0 0

Here we see that the condition is not true and it is in 0 0 form

So we need to further simply it to get standard form:

lim z→ f( z )= lim z→1 ( z 1 3 −1 ) z−1 ( z 1 6 −1 ) z−1 (On dividing num and den by the term z−1 )

On separating the limits we get,

lim z→1 ( z 1 3 −1 ) z−1 lim z→1 ( z 1 6 −1 ) z−1

According to theorem, for any positive integer n , lim x→a x n − a n x−a =n a n−1

So using above theorem and solving numerator and denominator separately, we get

lim z→1 ( z 1 3 −1 ) z−1 lim z→1 ( z 1 6 −1 ) z−1 = 1 3 ⋅ 1 −2 3 1 6 ⋅ 1 −5 6 = 1 3 1 6 =2

Thus, the value of the given expression lim z→1 ( z 1 3 −1 ) ( z 1 6 −1 ) =2 .