Limit Continuity and Differentiability 1 Question 3

4. limx0sin2x21+cosx equals

(2019 Main, 8 April I)

(a) 42

(b) 2

(c) 22

(d) 4

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Answer:

Correct Answer: 4. (a)

Solution:

  1. Given,

limx1+(1|x|+sin|1x|)sinπ2[1x]|1x|[1x]

Put x=1+h, then

x1+h0+

limx1+(1|x|+sin|1x|)sinπ2[1x]|1x|[1x]

=limh0+(1|h+1|+sin|h|)sinπ2[h]|h|[h]

=limh0+(1(h+1)+sinh)sinπ2[h]h[h]

(|h|=h and |h+1|=h+1 as h>0)

=limh0+(h+sinh)sinπ2(1)h(1)([x]=1 for 1<x<0 and h0+h0)=limh0+(h+sinh)hsinπ2=limh0+sinhhlimh0+hh=limh0+sinhhlimh0+hh=11=0limh0+sinhh=1



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