Problem 450 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } \lim _{x \rightarrow 0} \frac{\sinh 3 x+a \sinh 2 x+b \sinh x}{x^{5}} \text { is finite, find } a, b \text { and the value of the limit. } $$

Step-by-Step Solution

Answer

The values of \(a\) and \(b\) are \(-3/2\) and \(3/2\) respectively and the value of the limit as \(x \rightarrow 0\) is \(\frac{1}{20}\)

Step 1: Expansion using Maclaurin Series

Expand the functions in the numerator using the Maclaurin series \(\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots\) Our numerator then becomes \(\sinh 3x + a \sinh 2x + b \sinh x = 3x + \frac{9x^3}{3!} + \frac{81x^5}{5!} + \ldots + 2ax + \frac{4ax^3}{3!} + \frac{16ax^5}{5!} + \ldots + bx + \frac{bx^3}{3!} + \frac{bx^5}{5!} + \ldots\). This can be simplified to \(x(3 + 2a + b) + x^3 (\frac{9}{3!} + \frac{4a}{3!} + \frac{b}{3!}) + x^5(\frac{81}{5!} + \frac{16a}{5!} + \frac{b}{5!}) + \ldots\)

Step 2: Finding the values of \(a\) and \(b\)

For the limit to be finite when \(x \rightarrow 0\), the terms containing \(x\) and \(x^3\) in the numerator should cancel out, i.e., the coefficients of \(x\) and \(x^3\) must be 0. This gives us two equations: \(3 + 2a + b = 0\) and \(\frac{9}{2} + 2a + b = 0\). Solving these equations simultaneously, we find that \(a = -3/2\) and \(b = 3/2\)

Step 3: Finding the value of the limit

Inserting the obtained values of \(a\) and \(b\) into the function and simplifying shows that the limit as \(x \rightarrow 0\) is \(\frac{1}{20}\)

Key Concepts

Maclaurin Series

The Maclaurin series is a powerful tool for approximating functions using polynomials. It's essentially a Taylor series expansion centered around zero. For a function to be represented by a Maclaurin series, it should be expressed in the form:

• Function value at zero
• First derivative at zero, multiplied by the input variable
• Second derivative divided by factorial of two, times the square of the input, and so on for higher derivatives

For the hyperbolic sine function, \[\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \]This means the series expands into odd powers of \(x\) since the function sleeves up only the odd factorial terms. This type of expansion helps simplify complex expressions, especially when dealing with small values approaching zero, as seen in limit problems. Understanding Maclaurin series makes it easier to handle expressions like hyperbolic sine because it provides a polynomial approximation, making calculus operations such as limits more straightforward to compute.

Hyperbolic Functions

Hyperbolic functions such as \(\sinh x\), \(\cosh x\), and \(\tanh x\) have many applications in mathematics and physics, similar in spirit to trigonometric functions. They often appear in problems involving geometry and calculus of hyperbolas.

• \(\sinh x\) and \(\cosh x\) are analogs of sine and cosine functions.
• They follow specific identities, like \(\cosh^2 x - \sinh^2 x = 1\).

These functions provide results essential for calculating areas and lengths in hyperbolic geometry, as well as finding solutions to certain differential equations.
The hyperbolic sine, in particular, is noted for its simple polynomial series expansion, which is very helpful when analyzing functions around zero using limits. Its derivatives are also straightforward, paralleling sine and cosine in terms of rules of differentiation.

Simultaneous Equations

Solving simultaneous equations is a fundamental algebraic technique, essential when dealing with two or more equations that share variables. In the context of this problem, simultaneous equations are used to find specific parameter values that make an expression meet certain criteria, like finiteness at a limit.
We had two conditions that arose from ensuring that the limit was finite:

• \(3 + 2a + b = 0\)
• \(\frac{9}{2} + 2a + b = 0\)

These equations describe lines in a coordinate plane, and solving them involves finding the intersection, which represents values compatible with both equations. Using techniques like substitution or elimination, we can determine the values: \(a = -\frac{3}{2}\) and \(b = \frac{3}{2}\), thus solving the equation and allowing the calculation of the limit.

Finite Limits

The concept of finite limits is crucial in calculus, as it helps evaluate function behavior as it approaches a particular point. A finite limit exists when a function f(x) approaches a specific value as x tends toward some point, like zero, without diverging to infinity or oscillating indefinitely.
In this exercise, our aim was to ensure the expression reached a finite result by canceling terms that might lead to undefined behaviors, like division by zero.

• The terms with \(x\) and \(x^3\) in the expansion needed canceling from the numerator.
• This criterion led us to simplifying the expression correctly, guaranteeing the limit could be resolved as \(x\) approaches zero.

Using the Maclaurin series and results from solving simultaneous equations, we managed to rework the expression, ultimately confirming that the limit indeed existed and equaled \(\frac{1}{20}\). This process exemplifies the importance of limits in understanding precise point behaviors in calculus.