Problem 18 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

Values of constant \(a, b\) and \(c\) so that \(\lim _{x \rightarrow 0} \frac{x a e^{x}-b \log (1+x)+c x e^{-x}}{x^{2} \sin x}=2\), then \(a=\), \(b=, c=\) (a) \(3,12,9\) (b) \(9,6,9\) (c) \(5,10,20\) (d) None of these

Step-by-Step Solution

Answer

None of these (option d).

Step 1: Setup the Problem

We need to find the values of constants \(a\), \(b\), and \(c\) that satisfy the limit condition: \(\lim _{x \rightarrow 0} \frac{x a e^{x}-b \log (1+x)+c x e^{-x}}{x^{2} \sin x}=2\). Our goal is to analyze and simplify the expressions involved.

Step 2: Expand the Expressions

We'll expand each component of the expression in the numerator around \(x = 0\) using Taylor series:1. For \(e^x\), its expansion is \(1 + x + \frac{x^2}{2} + \ldots\).2. For \(\log (1 + x)\), its expansion is \(x - \frac{x^2}{2} + \ldots\).3. For \(e^{-x}\), its expansion is \(1 - x + \frac{x^2}{2} + \ldots\).4. For \(\sin x\), its expansion is \(x - \frac{x^3}{6} + \ldots\).

Step 3: Substitute and Simplify

Substitute the Taylor series expansions into the original expression:\[ x a (1 + x + \frac{x^2}{2}) - b (x - \frac{x^2}{2}) + c x (1 - x + \frac{x^2}{2}) \]= \[ ax + ax^2 + \frac{ax^3}{2} - bx + \frac{bx^2}{2} + cx - cx^2 + \frac{cx^3}{2} \]= \[ (a-b+c)x + (a + \frac{b}{2} - c)x^2 + \frac{a+c}{2}x^3 \].Substitute this into \(x^2 \sin x \approx x^3 - \frac{x^5}{6}\) and simplify.

Step 4: Solve for Constants

Equate the simplified expression \(a\) and \((a + \frac{b}{2} - c)\) terms to zero and find:1. The term \((a-b+c)\) must be zero for non-constant terms to vanish as \(x \to 0\).2. Set the limit equal to 2 by matching coefficients of \(x^2\) terms in the numerator. Solve: - \(a - b + c = 0\) - \(a + \frac{b}{2} - c = 2\).Thus, solve the system for \( a, b, c \): - Equation 1: \(a - b + c = 0\), - Equation 2: \(a + \frac{b}{2} - c = 2\).

Step 5: Check Each Option

Test each set of values from the options given:- Option (a): \(a = 3, b = 12, c = 9\) doesn't satisfy the second equation.- Option (b): \(a = 9, b = 6, c = 9\): - First equation: \(9 - 6 + 9 = 12 eq 0\).- Option (c): \(a = 5, b = 10, c = 20\): - Substitute into equations: \(5 - 10 + 20 = 15 eq 0\).None of the given options satisfy both conditions, so the answer is within option (d).

Key Concepts

Taylor Series Expansion

The Taylor series expansion is a powerful tool in calculus that allows us to express complicated functions as infinite sums of terms calculated from the values of that function's derivatives at a single point. By expanding functions like \({e^x}\), \({\log(1 + x)}\), and \(\sin x\) into series around a point (usually zero), we can approximate these functions with polynomials.
For example, the Taylor series for \({e^x}\) centered at zero is \({1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots}\), which provides a polynomial approximation.

• \({e^{-x}}\) expands to \(1 - x + \frac{x^2}{2} + \ldots\).
• \({\log(1+x)}\) can be approximated as \(x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots.\)
• \({\sin x}\) simplifies to \(x - \frac{x^3}{6} + \frac{x^5}{120} - \ldots.\)

When dealing with limits, Taylor series help simplify expressions by reducing them to manageable terms. This makes it easier to analyze their behavior as \(x\to0\). Such series are especially useful when working with indeterminate forms, simplifying calculations by focusing on polynomial terms.

L'Hôpital's Rule

L'Hôpital's Rule is a method that simplifies the evaluation of limits involving indeterminate forms such as \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). This rule states that, if the limits of both the numerator and the denominator approach 0 or infinity, the limit of their ratio can be found using the derivatives of these functions.
In practice, applying L'Hôpital's Rule involves these steps:

• Ensure that the initial limit you're dealing with is of the proper indeterminate form, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Compute the derivative of both the numerator and the denominator.
• Substitute back into the limit expression and evaluate again.

This approach often simplifies the form considerably, allowing you to find the limit. In the given exercise, even though L'Hôpital's Rule was not directly applied, the concept of indeterminate limits necessitated simplification and checking of terms, which is often remedied by using the rule. In instances where complex expressions don't resolve after a single application, multiple applications may be required.

Trigonometric Limits

Trigonometric limits form a crucial category of limits often encountered in calculus, especially important for calculations involving small angles. A commonly used property is that as \(x\to0\), \({\sin x}\to x\), which means that their ratio approaches 1, i.e., \({\frac{\sin x}{x}}\to 1\).
Such limits help simplify expressions in calculus problems. For example, when evaluating limits with functions involving \({\sin x}\), like the one present in the original exercise, it's typical to use approximations based on the Taylor series:

• \({\sin x}\ \approx x - \frac{x^3}{6} + \ldots\)
• As a result, in limits, replacing \({\sin x}\) by its approximation aids in factoring and simplifying the expression.

This kind of transformation is necessary when the function in question forms an indeterminate type. By substituting expansions, you can focus on the leading term to determine the function's behavior as \({x\to0}\). In doing so, you can match coefficients and solve for variables with ease.