Let `alpha` be the repeated root of `px^(2)+qx+r=0`, then `lim_(xrarralpha)(sin(px^(2)+qx+r))/((x-alpha)^(2))` equals

Options

A. 0
B. q
C. r
D. p

Answer

To solve the limit problem, we start with the expression:

\[ \lim_{x \to \alpha} \frac{\sin(px^2 + qx + r)}{(x - \alpha)^2} \]

Given that \(\alpha\) is a repeated root of the quadratic equation \(px^2 + qx + r = 0\), we know that:

\[ p\alpha^2 + q\alpha + r = 0 \]

This means that as \(x\) approaches \(\alpha\), \(px^2 + qx + r\) approaches 0. Therefore, we can rewrite our limit:

1. **Substituting \(\alpha\)**:
\[ \lim_{x \to \alpha} \frac{\sin(px^2 + qx + r)}{(x - \alpha)^2} = \frac{\sin(0)}{(x - \alpha)^2} = \frac{0}{0} \]
This is an indeterminate form \(0/0\).

2. **Using L'Hôpital's Rule**:
Since we have a \(0/0\) form, we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator:
- The derivative of the numerator \(\sin(px^2 + qx + r)\) is:
\[ \cos(px^2 + qx + r) \cdot (2px + q) \]
- The derivative of the denominator \((x - \alpha)^2\) is:
\[ 2(x - \alpha) \]

Thus, we can rewrite the limit as:
\[ \lim_{x \to \alpha} \frac{\cos(px^2 + qx + r) \cdot (2px + q)}{2(x - \alpha)} \]

3. **Evaluating the limit again**:
As \(x\) approaches \(\alpha\), \(px^2 + qx + r\) approaches \(0\), so \(\cos(px^2 + qx + r)\) approaches \(\cos(0) = 1\). Therefore, we have:
\[ \lim_{x \to \alpha} \frac{1 \cdot (2p\alpha + q)}{2(x - \alpha)} \]

This still results in an indeterminate form \(0/0\), so we apply L'Hôpital's Rule again.

4. **Second application of L'Hôpital's Rule**:
- The derivative of the numerator \(2p\alpha + q\) is \(0\) since it is constant.
- The derivative of the denominator \(2(x - \alpha)\) is \(2\).

Thus, we have:
\[ \lim_{x \to \alpha} \frac{0}{2} = 0 \]

5. **Final evaluation**:
We realize that we need to go back to our original limit and consider the Taylor expansion of \(\sin(u)\) around \(u = 0\):
\[ \sin(u) \approx u \quad \text{for small } u \]

Therefore, we can express:
\[ \sin(px^2 + qx + r) \approx px^2 + qx + r \]

So, substituting back into the limit:
\[ \lim_{x \to \alpha} \frac{px^2 + qx + r}{(x - \alpha)^2} \]

Since \(px^2 + qx + r = p(x - \alpha)^2\) (because \(\alpha\) is a repeated root), we have:
\[ \lim_{x \to \alpha} \frac{p(x - \alpha)^2}{(x - \alpha)^2} = p \]

Thus, the final answer is:

\[ \boxed{p} \]