Limits, Continuity & Differentiability
Jump Discontinuity
Grade 12
Question:
<p>Function whose jump (non-negative difference of LHL and RHL) of discontinuity is greater than or equal to one is/are</p><p>(a) $f(x) = \begin{cases} \frac{e^{1/x} + 1}{e^{1/x} - 1} & ; x \neq 0 \\ -\frac{1 + \cos x}{(x+1)^2} & ; 1 < x < 1 \end{cases}$</p><p>(b) $g(x) = \begin{cases} \frac{x^{1/3} + 1}{x^{1/2} - 1} & ; x \neq 0 \\ \ln x & ; 1 < x < 1 \end{cases}$</p><p>(c) $u(x) = \frac{\sin^{-1}(2x) - \frac{\pi}{14}}{3x}$ for $x \neq 0$; $u(0, 2) = 0, 2$</p><p>(d) $v(x) = \begin{cases} \log_3(x^2 + 2) & ; x < 2 \\ \tan\left(\frac{x}{2}\right) & ; 0 < x < 0 \\ \log_{1/2}(x + 5) & ; x > 2 \end{cases}$</p>
<p>(a) $f(x)$ as given</p>
<p>(b) $g(x)$ as given</p>
<p>(c) $u(x)$ as given</p>
<p>(d) $v(x)$ as given</p>
Step-by-Step Solution
Key Concept: Jump discontinuity is quantified by the absolute difference between right and left limits. Calculate LHL and RHL at suspected discontinuity points.
<p>Jump discontinuity occurs when $|\text{LHL} - \text{RHL}| \geq 1$ at a point.</p><p><strong>(a)</strong> At $x = 0$: $\lim_{x \to 0^-} \frac{e^{1/x} + 1}{e^{1/x} - 1} = \frac{0+1}{0-1} = -1$ and $\lim_{x \to 0^+} \frac{e^{1/x} + 1}{e^{1/x} - 1} = \frac{\infty + 1}{\infty - 1} = 1$. Jump $= |1 - (-1)| = 2 \geq 1$. ✓</p><p><strong>(b)</strong> Analysis of limits at $x=0$ shows jump $< 1$. ✗</p><p><strong>(d)</strong> Analyzing jump at $x = 2$ for the piecewise function yields jump $\geq 1$. ✓</p>
Correct Answer: A, D