Question

If \((\mathrm{a} / 2)\) and \((\mathrm{b} / 2)\) be two distinct real roots of \(\ell \mathrm{x}^{2}+\mathrm{mx}+\mathrm{n}=0\) then \(\lim _{\mathrm{x} \rightarrow(\mathrm{a} / 2)}\left[\left\\{1-\operatorname{Cos}\left(\ell \mathrm{x}^{2}+\mathrm{mx}+\mathrm{n}\right)\right\\} /(2 \mathrm{x}-\mathrm{a})^{2}\right]=?\) (Where \(\ell=0, \mathrm{a}, \mathrm{b} \in \mathrm{R})\) (a) \(\left[\ell^{2} /\left\\{8(\mathrm{a}-\mathrm{b})^{2}\right\\}\right]\) (b) \(\left(\ell^{2} / 32\right)\left(a^{2}-b^{2}\right)\) (c) \(\left(\ell^{2} / 32\right)(\mathrm{a}-\mathrm{b})^{2}\) (d) \(\left(\ell^{2} / 16\right)\left(a^{2}-b^{2}\right)\)

Short answer

The short answer is: \[\lim_{x\to\frac{a}{2}}\frac{1 - \cos(\ell x^2 + mx + n)}{(2x-a)^2} = \boxed{\frac{\ell^2}{16}(a^2-b^2)}\]

Step-by-step solution

Rewrite the function using the roots

Since the quadratic equation has roots \(\frac{a}{2}\) and \(\frac{b}{2}\), we can rewrite it as follows: \[ \ell x^2 + mx + n = \ell (x - \frac{a}{2})(x - \frac{b}{2}) \]

Find the expression for the quadratic equation

Expand the factored form of the quadratic equation from step 1 to obtain the expression for the quadratic equation: \[ \ell x^2 + mx + n = \ell\left(x^2 - (\frac{a + b}{2})x + \frac{a b}{4}\right) \]

Apply the limit properties

Recall the limit of the product is the product of the limits when the individual limits exist. Apply this property to the given expression: \[ \lim_{x\to\frac{a}{2}}\frac{1 - \cos(\ell x^2 + mx + n)}{(2x-a)^2} = \left(\lim_{x\to\frac{a}{2}}\frac{1 - \cos(\ell x^2 + mx + n)}{\ell x^2 + mx + n}\right)\times\left(\lim_{x\to\frac{a}{2}}\frac{\ell x^2 + mx + n}{(2x-a)^2}\right) \]

Apply L'Hopital's Rule

Now use L'Hopital's Rule on the left limit. The derivative of the numerator is: \[ \frac{d}{dx}\,(1 - \cos(\ell x^2 + mx + n)) = 2\ell x\sin(\ell x^2 + mx + n) + m\sin(\ell x^2 + mx + n) \] The derivative of the denominator is: \[ \frac{d}{dx}\, (\ell x^2 + mx + n) = 2\ell x + m \] Now, apply L'Hopital's Rule: \[ \lim_{x\to\frac{a}{2}} \frac{1-\cos(\ell x^2 + mx + n)}{\ell x^2 + mx + n} = \lim_{x\to\frac{a}{2}}\frac{2\ell x\sin(\ell x^2 + mx + n) + m\sin(\ell x^2 + mx + n)}{2\ell x + m} \]

Simplify the limit expression

Evaluating the limit using the previous step results and properties of sine and cosine, we have: \[ \lim_{x\to\frac{a}{2}} \frac{2\ell x\sin(\ell x^2 + mx + n) + m\sin(\ell x^2 + mx + n)}{2\ell x + m} = \lim_{x\to\frac{a}{2}}\frac{2\ell x\cdot 1 + m\cdot1}{2\ell x + m} = 1 \]

Evaluate the second limit

Given that \(x\to \frac{a}{2}\): \[ f(x) = \ell x^2 + mx + n = \ell(x - \frac{a}{2})(x - \frac{b}{2}) \] Since \(x \to \frac{a}{2}\), the expression becomes: \[ f\left(\frac{a}{2}\right) = \ell\left(\frac{a}{2} - \frac{a}{2}\right)\left(\frac{a}{2} - \frac{b}{2}\right) = -\frac{\ell(a-b)^2}{4} \] Now, consider the limit \(\lim_{x\to\frac{a}{2}}\frac{\ell x^2 + mx + n}{(2x-a)^2}\): \[ \lim_{x\to\frac{a}{2}}\frac{\ell x^2 + mx + n}{(2x-a)^2} = \lim_{x\to\frac{a}{2}} \frac{-\frac{\ell(a-b)^2}{4}}{(2x-a)^2} \]

Combine the limits and simplify

Using the product of the limits from step 3: \[ \lim_{x\to\frac{a}{2}}\frac{1 - \cos(\ell x^2 + mx + n)}{(2x-a)^2} = 1 \cdot \lim_{x\to\frac{a}{2}} \frac{-\frac{\ell(a-b)^2}{4}}{(2x-a)^2} = \frac{-\ell(a-b)^2}{4\left(\frac{a^2}{4}\right)} = \boxed{\frac{\ell^2}{16}(a^2-b^2)} \] Thus, the correct answer is option (d).

Key concepts

Root of a Quadratic Equation

Understanding the root of a quadratic equation is foundational when handling more complex mathematical problems involving limits and functions. A quadratic equation typically takes the form \( ax^2 + bx + c = 0 \), where \( a \) is not equal to zero, and its roots are the values of \( x \) for which the equation holds true.

When it comes to identifying the roots, there are multiple methods, such as factoring, completing the square, or using the quadratic formula. In our example, knowing that \( \frac{a}{2} \) and \( \frac{b}{2} \) are roots of the given equation allows us to rewrite the equation by expressing it as \( \ell(x - \frac{a}{2})(x - \frac{b}{2}) = 0 \) and use this factorized form to simplify the process of finding limits. This insight provides a shortcut to otherwise lengthy substitutions and evaluations.

Applying this knowledge to the limit calculation helps us understand how the behavior of the function changes as it approaches the values of its roots, which is crucial when dealing with continuous functions and their properties on a graph.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool for calculating limits that appear to be indeterminate forms, like \( 0/0 \) or \( \infty/\infty \) . The rule states that if the limits of the numerator and the denominator of a fraction both approach zero or infinity, the limit of the fraction equals the limit of the derivatives of the numerator and the denominator.

Applying L'Hopital's Rule requires us to take derivatives. For our problem, differentiating the top and bottom separately and taking the limit is the key to finding the limit of the entire function as \( x \) approaches \( \frac{a}{2} \). This is particularly useful when working with trigonometric functions within limits, as it simplifies complex expressions into manageable forms.

By using L'Hopital's Rule in the exercise, we are able to transform an indeterminate form into a limit that we can evaluate directly. This demonstrates the rule's utility in resolving what might seem like insurmountable mathematical hurdles and is essential for any student tackling calculus problems involving limits.

Trigonometric Limits

Trigonometric limits are an essential aspect of calculus, particularly when we look at the behavior of functions involving sine, cosine, and other trigonometric ratios as the input approaches a certain value. The challenge lies in the oscillating nature of these functions, which makes finding their limits less straightforward.

In the given exercise, we encounter a limit involving the cosine function. Trigonometric identities and properties often come into play, making it possible to simplify expressions before evaluating the limit. For example, knowing that \( 1 - \cos(x) \) approaches zero as \( x \) approaches zero can help evaluate limits without actual substitution.

It’s also important to recognize when a trigonometric limit presents an indeterminate form, as in this case with \( 1 - \cos(\ell x^2 + mx + n) \) over \( (2x-a)^2 \) as \( x \) approaches \( \frac{a}{2} \). Here, we utilize L'Hopital's Rule to circumvent the indeterminate form and find the exact value of the limit. Establishing a strong understanding of trigonometric limits is crucial for students to effectively navigate the intricate waters of calculus.