Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>Let <span>\( k \)</span> be a non-zero real number. If <span>\[ f(x) = \begin{cases} \dfrac{(e^x - 1)^2}{\sin\!\left(\dfrac{x}{k}\right)\log\!\left(1 + \dfrac{x}{4}\right)}, & x \neq 0 \\ 12, & x = 0 \end{cases} \]</span> is a continuous function, then the value of <span>\( k \)</span> is</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, the limit of the piecewise function must equal f(0)=12. Use standard limits: (e^x-1)~x, sin(x/k)~x/k, and log(1+x/4)~x/4 as x→0 to find k.
<p><strong>Step 1:</strong> For continuity at x=0, we need: $\lim_{x \to 0} f(x) = f(0) = 12$</p><p><strong>Step 2:</strong> Apply standard limits as $x \to 0$:<br>• $(e^x - 1)^2 \sim x^2$<br>• $\sin\left(\frac{x}{k}\right) \sim \frac{x}{k}$<br>• $\log\left(1 + \frac{x}{4}\right) \sim \frac{x}{4}$</p><p><strong>Step 3:</strong> Substitute into the limit:<br>$$\lim_{x \to 0} \frac{(e^x-1)^2}{\sin(x/k)\log(1+x/4)} = \lim_{x \to 0} \frac{x^2}{\frac{x}{k} \cdot \frac{x}{4}}$$</p><p><strong>Step 4:</strong> Simplify:<br>$$= \lim_{x \to 0} \frac{x^2}{\frac{x^2}{4k}} = \lim_{x \to 0} \frac{x^2 \cdot 4k}{x^2} = 4k$$</p><p><strong>Step 5:</strong> For continuity: $4k = 12$<br>$$k = 3$$</p><p>∴ Answer: <strong>k = 3</strong> (Option C)</p>
Correct Answer: C