Limit Continuity and Differentiability 4 Question 4

4. Letf(x)=limnnn(x+n)x+n2x+nnn!(x2+n2)x2+n24x2+n2n2,

for all x=0. Then

(a) f12f(1)

(b) f13f23

(c) f(2)0

(d) f(3)f(3)f(2)f(2)

(2016 Adv.)

Numerical Value

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Answer:

Correct Answer: 4. (c)

Solution:

  1. Given function f(x)=[x]x4,xR

Now, limx4+f(x)=limh0[4+h]4+h4

[ put x=4+h, when x4+, then h0 ]

=limh0(41)=3

and limx4f(x)=limh0[4h]4h4

[ put x=4h, when x4

then h0]

=limh0(30)=3

and f(4)=[4]44=41=3

limx4f(x)=f(4)=limx4+f(x)=3

So, function f(x) is continuous at x=4.



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