Question

If f (x)$=\{\frac{|x+2|}{ta{n}^{-1}{(}_{2}x+2)}x\ne -2$
$x=-2,$ then f(x) is

• A. continuous at $x=-2$
• B. not continuous at $x=-2$
• C. differentiable at $x=-2$
• D. continuous but not differentiable at $x=-2$

Answer 1

Answer: (B)

not continuous at $x=-2$

L.H.L$={lim}_{x\to -2^{-}}f\left(x\right)={lim}_{h\to 0^{+}}f(-2-h)$
$={lim}_{h\to 0^{+}}\frac{|-2-h+2|}{ta{n}^{-1}(-2-h+2)}$
${lim}_{h\to 0^{+}}\frac{h}{ta{n}^{-1}(-h)}$
${lim}_{h\to 0^{+}}\frac{h}{-ta{n}^{-1}h}$
& R.H.L. $={lim}_{x\to -2^{+}}f\left(x\right)={lim}_{h\to 0^{+}}{f}_{-}2+h$
$={lim}_{h\to 0^{+}}\frac{|-2+h+2|}{ta{n}^{-1}(-2+h+2)}$
$={lim}_{h\to 0^{+}}\frac{h}{ta{n}^{-1}\left(h\right)}=1$
$\therefore {lim}_{x\to -2^{-}}f\left(x\right)\ne {lim}_{x\to -2^{+}}f\left(x\right)$
so, f(x) is not continuous & not differentiable at x $=-2$