Limits, Continuity & Differentiability
Continuity
Grade 12
Question:
<p>If \[f(x) = \begin{cases} \dfrac{\sin(p+1)x + \sin x}{x}, & x < 0 \\ q, & x = 0 \\ \dfrac{\sqrt{x+x^2} - \sqrt{x}}{x^{3/2}}, & x > 0 \end{cases}\] is continuous at \(x = 0\), then the ordered pair \((p,\,q)\) is equal to:</p>
<p>\(\left(-\dfrac{3}{2},\,-\dfrac{1}{2}\right)\)</p>
<p>\(\left(-\dfrac{1}{2},\,\dfrac{3}{2}\right)\)</p>
<p>\(\left(-\dfrac{3}{2},\,\dfrac{1}{2}\right)\)</p>
<p>\(\left(\dfrac{5}{2},\,\dfrac{1}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, the left limit, right limit, and f(0) must all be equal. Use the standard limit lim(x→0) sin(x)/x = 1 and L'Hôpital's rule or series expansion to evaluate the limits from both sides.
**Step 1:** For the function $f(x)$ to be continuous at $x=0$, the limit of $f(x)$ as $x$ approaches $0$ must exist and be equal to the function's value at $x=0$.
$$ \lim_{x \to 0} f(x) = f(0) $$
**Step 2:** Calculate the limit of $f(x)$ as $x \to 0$.
For $x \neq 0$, the function is defined as $f(x) = \dfrac{\sin(p+1)x + \sin x}{x}$.
$$ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(p+1)x + \sin x}{x} $$
We can split the fraction into two terms:
$$ = \lim_{x \to 0} \left( \frac{\sin(p+1)x}{x} + \frac{\sin x}{x} \right) $$
To evaluate the first term, we multiply and divide by $(p+1)$:
$$ = \lim_{x \to 0} \left( (p+1) \frac{\sin(p+1)x}{(p+1)x} + \frac{\sin x}{x} \right) $$
Using the standard limit $\lim_{u \to 0} \frac{\sin u}{u} = 1$:
$$ = (p+1)(1) + 1 $$
$$ = p+1+1 = p+2 $$
**Step 3:** Equate the limit to $f(0)$.
Given $f(0) = q$.
From the condition for continuity (Step 1) and the calculated limit (Step 2), we must have:
$$ q = p+2 $$
Thus, the ordered pair $(p,q)$ must satisfy the relationship $q = p+2$.
Correct Answer: C