Question

If f(x)={|x|+1, x <0 0, x=0 |x|-1, x>0}
For what value (s) of a does lim x\(\rightarrow\)af(x) exist?

Answer 1

The given function is f(x)= {|x| +1, x<0 0, x=0 |x| +1, x-1, x>0
When a = 0,
\(\lim_{x\rightarrow 0^-}\) f(x)= lim x\(\rightarrow\)0- (|x|+1)
\(\lim_{x\rightarrow 0^-}\)(-x+1) [if x<0m |x| = -x ]
=-0+1
=1
\(\lim_{x\rightarrow 0^+}\)f(x)= lim x\(\rightarrow\)0⁺ (|x|-1)
=\(\lim_{x\rightarrow 0^+}\) (x-1) [If x > 0, |x| = x]
=0-1
=-1
Here, it is observed that lim x\(\rightarrow\)0-ƒ (x) ≠ \(\lim_{x\rightarrow 0^+}\) ƒ (x).
∴lim f(x) does not exist.
When a <0,
\(\lim_{x\rightarrow 0^-}\) f(x)= lim (|x|+1)
=\(\lim_{x\rightarrow a}\)(-x+1) [x<a<0 \(\Rightarrow\) |x|= -x]
=-a+1
\(\lim_{x\rightarrow a^+}\)f(x)= \(\lim_{x\rightarrow a^+}\)(|x|+1)
=\(\lim_{x\rightarrow a}\)(-x+1) [a<x<0 \(\Rightarrow\) |x|=-x]
=-a+1
∴\(\lim_{x\rightarrow a^-}\) f(x)= \(\lim_{x\rightarrow a^+}\) f(x)=−a+1 
Thus, the limit of f (x) exists at x = a, where a < 0.
When a > 0
\(\lim_{x\rightarrow a^-}\) f(x)= \(\lim_{x\rightarrow a^-}\)(|x|-1)
= \(\lim_{x\rightarrow a}\) (x-1) [0<x<a \(\Rightarrow\) |x|=x]
=a-1
\(\lim_{x\rightarrow a^+}\) f(x)= \(\lim_{x\rightarrow a^+}\)(|x|-1)
= \(\lim_{x\rightarrow a}\) (x-1) [0<a<x \(\Rightarrow\) |x|=x]
=a-1
∴\(\lim_{x\rightarrow a^-}\) f(x)= \(\lim_{x\rightarrow a^+}\) f(x)=a-1
Thus, limit of f (x)exists at x = a, where a > 0.
Thus, lim x\(\rightarrow\)a f(x) exists for all a ≠ 0.