Diff calc... limits....

Hey frnds.... plz help me out here........

i tried to do in a graphical method.... but can't succeed.....

f(x) = [x] + 1 , x≠nπ ; n= 0 , + - 1 , + -2,......
= 3 otherwise

g(x) = x2+1 ; x≠3,x≠0
3 ; x=0
5 ; x=3

then... Lim (x→0) g(f(x)) is.....

plz try it out......

3 Answers

1
nihal raj ·

when x=0 f(x)=3 & lim(x---0) g(f(x))=g(3)=5

when x=3 f(x)=[x]+1=3+1=4 & lim(x--0) g(f(x))=(4)2+1=17

when x=n f(x)=3 & lim(x--0) g(f(x))=5

that is the answer

1
Euclid ·

hey nihal..... hw cud u get f(x)= [x] +1 wen x=3??? and y did u consider x=3???
plzzz xplain

i think ans is 1

wen x-->0 (not exactly 0) f(x) = [x]+1

1
nihal raj ·

FIRSTLY U SHOULD KNOW THAT U HAVE TO CALCULATE LIMX---0 g(f(x)) and not LIMX---0f(x).. answer is................ LIMX---0 g(f(x)) =

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