If f(x) = begin{cases} |x|+1, & x

Question

(A) The given function is $f(x) = \begin{cases} |x|+1, & x  0 \end{cases}$.Case $1$: $a = 0$.$\lim_{x 	o 0^-} f(x) =

Vedclass · Accepted Answer

$a 
eq 0$

Vedclass · Answer

(A) The given function is $f(x) = \begin{cases} |x|+1, & x  0 \end{cases}$.Case $1$: $a = 0$.$\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (-x+1) = 1$.$\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (x-1) = -1$.Since $1 
eq -1$,the limit does not exist at $a = 0$.Case $2$: $a $\lim_{x 	o a^-} f(x) = \lim_{x 	o a^-} (-x+1) = -a+1$.$\lim_{x 	o a^+} f(x) = \lim_{x 	o a^+} (-x+1) = -a+1$.Since the left-hand limit equals the right-hand limit,the limit exists for all $a Case $3$: $a > 0$.$\lim_{x 	o a^-} f(x) = \lim_{x 	o a^-} (x-1) = a-1$.$\lim_{x 	o a^+} f(x) = \lim_{x 	o a^+} (x-1) = a-1$.Since the left-hand limit equals the right-hand limit,the limit exists for all $a > 0$.Conclusion: The limit exists for all $a 
eq 0$.

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

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