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.