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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 10
Total: 72
lim
z→1
z
1
3
−1
z
1
6
−1
Solution:  
We have ,
lim
z→1
z
1
3
−1
z
1
6
−1
 (0/0 form)
=
lim
z→1
(z
1
6
)2−1
(z
1
6
−1)
 =
lim
z→1
(z
1
6
−1)(z
1
6
+1)
(z
1
6
−1)
=
lim
z→1
(z
1
6
+1) = 1
1
6
 + 1 = 1 + 1 = 2.
Alternative solution :
lim
z→1
z
1
3
−1
z
1
6
−1
 =
lim
z→1
z
1
3
−1
z−1
×
z−1
z
1
6
−1
=
1
3
(1)(
1
3
)−1 ×
1
(
1
6
)(1)(
1
6
)−1
=
1
3
×
6
1
 = 2
Using
lim
x→a
xn−an
x−a
 = nan−1
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