If `alpha, beta` are the zeroes of `ax^(2)+bx+c`, then evaluate :
`lim_(x to beta)(1-cos(ax^(2)+bx+c))/(x-beta)^(2)`.

Answer

To evaluate the limit

\[ \lim_{x \to \beta} \frac{1 - \cos(ax^2 + bx + c)}{(x - \beta)^2} \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic polynomial \(ax^2 + bx + c\), we can follow these steps:

### Step 1: Substitute the expression for \(ax^2 + bx + c\)
Since \(\alpha\) and \(\beta\) are the roots of the polynomial, we can express \(ax^2 + bx + c\) in terms of its roots:

\[ ax^2 + bx + c = a(x - \alpha)(x - \beta) \]

### Step 2: Rewrite the limit
Now, we can rewrite the limit using this expression:

\[ \lim_{x \to \beta} \frac{1 - \cos(a(x - \alpha)(x - \beta))}{(x - \beta)^2} \]

### Step 3: Use the trigonometric identity
We know that \(1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)\). Thus, we can rewrite the limit as:

\[ \lim_{x \to \beta} \frac{2\sin^2\left(\frac{a(x - \alpha)(x - \beta)}{2}\right)}{(x - \beta)^2} \]

### Step 4: Change of variable
Let \(h = x - \beta\). As \(x \to \beta\), \(h \to 0\). Therefore, we can rewrite the limit in terms of \(h\):

\[ \lim_{h \to 0} \frac{2\sin^2\left(\frac{a((h + \beta) - \alpha)h}{2}\right)}{h^2} \]

### Step 5: Simplify the expression inside the sine function
Now, we can simplify the expression inside the sine function:

\[ = \lim_{h \to 0} \frac{2\sin^2\left(\frac{a(h + \beta - \alpha)h}{2}\right)}{h^2} \]

### Step 6: Use the small angle approximation
As \(h \to 0\), we can use the approximation \(\sin(x) \approx x\) for small \(x\):

\[ \sin\left(\frac{a(h + \beta - \alpha)h}{2}\right) \approx \frac{a(h + \beta - \alpha)h}{2} \]

### Step 7: Substitute back into the limit
Substituting this approximation back into the limit gives:

\[ \lim_{h \to 0} \frac{2\left(\frac{a(h + \beta - \alpha)h}{2}\right)^2}{h^2} \]

### Step 8: Simplify the limit
This simplifies to:

\[ \lim_{h \to 0} \frac{a^2(h + \beta - \alpha)^2h^2}{2h^2} = \lim_{h \to 0} \frac{a^2(h + \beta - \alpha)^2}{2} \]

### Step 9: Evaluate the limit
As \(h \to 0\), \(h + \beta - \alpha \to \beta - \alpha\):

\[ = \frac{a^2(\beta - \alpha)^2}{2} \]

Thus, the final answer is:

\[ \lim_{x \to \beta} \frac{1 - \cos(ax^2 + bx + c)}{(x - \beta)^2} = \frac{a^2(\beta - \alpha)^2}{2} \]