Limits, Continuity & Differentiability
Continuity analysis of piecewise functions
Grade 12
Question:
<p>On the interval <span>\(I = [-2, 2]\)</span>, if the function <span>\(f(x) = \begin{cases} (x^2 + 1)e^{-\frac{1}{|x|}} + \frac{1}{x} & , x \neq 0 \\ 0 & , x = 0 \end{cases}\)</span> then which of the following hold good?</p>
<p>(a) <span>\(f(x)\)</span> is continuous for all values of <span>\(x \in I\)</span></p>
<p>(b) <span>\(f(x)\)</span> is continuous for <span>\(x \in I \setminus \{0\}\)</span></p>
<p>(c) <span>\(f(x)\)</span> assumes all intermediate values from <span>\(f(-2)\)</span> to <span>\(f(2)\)</span></p>
<p>(d) <span>\(f(x)\)</span> has a maximum value equal to <span>\(3/e\)</span></p>
Step-by-Step Solution
Key Concept: Check continuity at points where the function definition changes or where denominators vanish; the behavior of exponential and reciprocal terms near special points determines overall continuity.
<p><strong>Analysis:</strong> Consider the function at <span>$x = 0$</span>. We have <span>$\lim_{x \to 0} (x^2 + 1)e^{-\frac{1}{|x|}} = 1 \cdot 0 = 0$</span> since the exponential term dominates. However, <span>$\lim_{x \to 0} \frac{1}{x}$</span> does not exist (left and right limits differ), so the function is discontinuous at <span>$x = 0$</span>.</p><p>For <span>$x \in I \setminus \{0\}$</span>, the function is continuous as it is a sum and product of continuous functions.</p><p>∴ Answer is (b).</p>
Correct Answer: B