Limits, Continuity & Differentiability
Continuity of piecewise functions
Grade 12
Question:
<p>Let <span>\(f(x)=\begin{cases}\frac{a[x]+x-1}{[x]+x}, & x\neq 0\\\log_e a, & x=0\end{cases}\)</span> where <span>\(a>0\)</span>. The function is</p>
<p>(a) continuous</p>
<p>(b) discontinuous</p>
<p>(c) cannot be determined</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, both one-sided limits must equal f(0) and each other. The behavior of [x] near 0 is different from the left and right, causing the limits to differ.
<p><strong>Step 1:</strong> Find the right-hand limit (RHL) as <span>$x\to 0^+$</span>.</p><p>For small positive <span>$h$</span>, <span>$[h]=0$</span>.</p><p><span>$\text{RHL}=\lim_{h\to 0^+}\frac{a[h]+h-1}{[h]+h}=\lim_{h\to 0^+}\frac{a\cdot 0+h-1}{0+h}=\lim_{h\to 0^+}\frac{h-1}{h}=\log_e a$</span></p><p><strong>Step 2:</strong> Find the left-hand limit (LHL) as <span>$x\to 0^-$</span>.</p><p>For small negative <span>$-h$</span>, <span>$[-h]=-1$</span>.</p><p><span>$\text{LHL}=\lim_{h\to 0^+}\frac{a[-h]-h-1}{[-h]-h}=\lim_{h\to 0^+}\frac{a(-1)-h-1}{-1-h}=\lim_{h\to 0^+}\frac{-a-h-1}{-1-h}=\frac{-a-1}{-1}=1-a$</span></p><p><strong>Step 3:</strong> Compare limits.</p><p><span>$\text{RHL}=\log_e a$</span> and <span>$\text{LHL}=1-a$</span>.</p><p>Since <span>$\log_e a\neq 1-a$</span> in general (RHL ≠ LHL), the function is discontinuous at <span>$x=0$</span>.</p><p><strong>∴ Answer is (b) discontinuous.</strong></p>
Correct Answer: B