If $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$, then $\lim _{x \rightarrow \alpha}…
If $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$, then $\lim _{x \rightarrow \alpha} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\alpha)^2}=$
$a^2(\alpha-\beta)^2$
$4 a^2(\alpha-\beta)^2$
$\frac{a^2}{2}(\alpha-\beta)^2$
$2 a^2(\alpha-\beta)^2$
Solution
Factorize the quadratic expression:
Since \(\alpha \) and \(\beta \) are the roots of the equation \(ax^{2}+bx+c=0\), the quadratic expression can be written as \(a(x-\alpha )(x-\beta )\). Substitute the factored form into the limit expression:
The given limit becomes \(\lim _{x\rightarrow \alpha }\frac{1-\cos \left(a(x-\alpha )(x-\beta )\right)}{(x-\alpha )^{2}}\). Apply the trigonometric limit identity: .rPeykc br{display:none}
It is known that \(\lim _{y\rightarrow 0}\frac{1-\cos y}{y^{2}}=\frac{1}{2}\). Let \(y=a(x-\alpha )(x-\beta )\). As \(x\rightarrow \alpha \), \(y\rightarrow a(\alpha -\alpha )(\alpha -\beta )=0\).
Rewrite the limit using the identity: The expression can be rewritten as \(\lim _{x\rightarrow \alpha }\frac{1-\cos \left(a(x-\alpha )(x-\beta )\right)}{\left(a(x-\alpha )(x-\beta )\right)^{2}}\cdot \frac{\left(a(x-\alpha )(x-\beta )\right)^{2}}{(x-\alpha )^{2}}\). Simplify the expression:
This simplifies to \(\lim _{x\rightarrow \alpha }\frac{1}{2}\cdot \frac{a^{2}(x-\alpha )^{2}(x-\beta )^{2}}{(x-\alpha )^{2}}\). Cancel out common terms and evaluate the limit: The \((x-\alpha )^{2}\) terms cancel, leaving \(\lim _{x\rightarrow \alpha }\frac{1}{2}a^{2}(x-\beta )^{2}\).
Substituting \(x=\alpha \), the limit is evaluated as \(\frac{1}{2}a^{2}(\alpha -\beta )^{2}\).
Final Answer
The final answer is \(\frac{a^{2}}{2}(\alpha -\beta )^{2}\)