Continuity and DifferentiabilityHard
Question
Options
A.exists and it equals √2
B.exists and it equals - √2
C.does not exist because x - 1 → 0
D.does not exist because left hand limit is not equal to right hand limit
Solution
LHL = 
Put x = 1- h, h > 0
for x → 1+, h → 0
= √2 
= - √2
Again, RHL
Put x = 1+ h, h > 0
for x → 1+, h → 0
∴ LHL ≠ RHL
Hence,
f(x) does not exist.
Put x = 1- h, h > 0
for x → 1+, h → 0
= √2 = - √2 Again, RHL
Put x = 1+ h, h > 0
for x → 1+, h → 0
∴ LHL ≠ RHL
Hence,
f(x) does not exist.Create a free account to view solution
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