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Question

If f(x)= |x|+1,x<00, x=0x3|x|1 x>0
For what value (s) of a does limxaf(x) exists?

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Solution

The given function is
f(x)= |x|+1,x<00, x=0x3|x|1 x>0
When a=0
limxaf(x)=limxa(|x|+1)
= limx0(x+1) [Ifx<0,|x|=x]
=0+1
=1
limxaf(x)=limxa+(|x|1)
= limx0(x1) [Ifx<0,|x|=x]
=01
=1
Here it is observed that limxaf(x)limxa+f(x)
limx0f(x) does not exist
When a<0
limxaf(x)=limxa(|x|+1)
= limxa(x+1) [x<a<0|x|=x]
=a+1
limxa+f(x)=limxa+(|x|+1)
limxa(x+1) [a<x<0|x|=x]
=a+1
limxaf(x)=limxa+f(x)=a+1
Thus limit of f(x) exists at x=a where a<0
When a>0
limxaf(x)=limxa(|x|1)
= limxa(x1) [0<x<a|x|=x]
=a1
limxa+f(x)=limxa+(|x|1)
= limxa(x1) [0<a<x|x|=x]
=a1
limxaf(x)=limxa+f(x)=a1
Thus limit of f(x) exists at x=a where a>0
Thus limxaf(x) exists for all a0

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