Question

If \[ f(x)= \begin{cases} \dfrac{x^{2}-4}{\sqrt{2-x}}, & x<2\\ a, & x=2\\ \log(x-2), & x>2 \end{cases} \] is a real valued function, then:

• A. f is continuous at \(x=2\) when \(a=0\)
• B. f is left continuous at \(x=2\) when \(a=0\)
• C. f is right continuous at \(x=2\) when \(a=\log 2\)
• D. f is not continuous at \(x=1\)

Hint:

A function is left continuous at \(x=a\) if \(\lim_{x\to a^-}f(x)=f(a)\), even if the right-hand limit does not exist.

Answer 1

The Correct Option is B

Step 1: Find the left-hand limit. \[ \lim_{x\to2^-} \frac{x^2-4}{\sqrt{2-x}} \] Factor: \[ = \lim_{x\to2^-} \frac{(x-2)(x+2)}{\sqrt{2-x}} \] Since \[ x-2=-(2-x) \] we get \[ = \lim_{x\to2^-} -\sqrt{2-x}(x+2) \] \[ =0 \] Hence, \[ \lim_{x\to2^-}f(x)=0 \]

Step 2: Find the right-hand limit. \[ \lim_{x\to2^+}\log(x-2) \] As \(x\to2^+\), \[ x-2\to0^+ \] Therefore, \[ \lim_{x\to2^+}\log(x-2) = -\infty \]

Step 3: Check continuity. Since \[ \lim_{x\to2^-}f(x)=0 \] and \[ \lim_{x\to2^+}f(x)=-\infty \] the two-sided limit does not exist. Therefore the function can never be continuous at \(x=2\). However, if \[ a=0 \] then \[ f(2)=0 = \lim_{x\to2^-}f(x) \] Thus the function is

left continuous at \(x=2\). \[ \boxed{\text{(B) f is left continuous at }x=2\text{ when }a=0} \]