Question

If \(\lim _{\theta \rightarrow 0}\left(\frac{1+a \cos \theta}{\theta^{2}}-\frac{b \sin \theta}{\theta^{3}}\right)=1\), then the value of \(\mathrm{a}+\mathrm{b}\) is \(-\lambda\), find the value of \(\lambda\). (1) 4 (2) 1 (3) 2 (4) 0

Short answer

The value of \lambda is 1.

Step-by-step solution

Simplify the limit expression

First, analyze each term separately. Use the standard limits, \[ \text{lim}_{\theta \to 0} \frac{\theta}{\theta} = 1 \] , \[ \text{lim}_{\theta \to 0} \frac{\theta^2}{\theta^2} = 1 \] and \[ \text{lim}_{\theta \to 0} \frac{\theta^3}{\theta^3} = 1 \] to simplify the given expression.-> Simplify the cosine term and sine term around \( \theta \to 0 \).

Apply Taylor Series Expansion

Use the Taylor series expansion for \( \theta \to 0 \) to approximate the trigonometric functions:\[ \text{cos}(\theta) \thickapprox 1 - \frac{\theta^2}{2} \]\[ \text{sin}(\theta) \thickapprox \theta - \frac{\theta^3}{6} \]Substitute these approximations into the given limit expression.

Substitute Taylor Expansions into the Limit

Substitute \( \text{cos}(\theta) \) and \( \text{sin}(\theta) \) from the Taylor Expansion:\[ \frac{1+a(1-\frac{\theta^2}{2})}{\theta^2} - \frac{b(\theta - \frac{\theta^3}{6})}{\theta^3} \]Simplify the expression step by step.

Simplify Each Fraction

Simplify the fractions from the substituted expansions:\[ \frac{1+a - a\frac{\theta^2}{2}}{\theta^2} - \frac{b\theta - b\frac{\theta^3}{6}}{\theta^3} \]Divide each term by \( \theta^2 \) and \( \theta^3 \) respectively:\[ \frac{1+a}{\theta^2} - \frac{a}{2} - \frac{b}{\theta} + \frac{b}{6} \]

Evaluate the Limit

Take the limit as \( \theta \to 0 \). Notice terms with higher powers of \( \theta \) in the denominator will tend towards infinity,\[ \text{lim}_{\theta \to 0}\frac{1+a}{\theta^2} - \frac{b}{\theta} = 1 \]To satisfy this limit, the only possible solution is both \( 1+a \) and \( b \) should be 0: \( a = -1 \) and \( b = 0 \).

Find \lambda

We know that \( a + b = -\lambda \). Substitute the values obtained: \( a = -1 \) and \( b = 0 \),\( -1 + 0 = -\lambda \).This gives us \( \lambda = 1 \).

Key concepts

Limits and Continuity

Limits and continuity are cornerstone concepts in calculus and essential for understanding more complex topics. First, let's talk about limits. A limit describes the value that a function approaches as the input approaches some value. For example, \(\text{lim}_{\theta \to 0} f(\theta)\) tells us what value f(\theta) is getting close to when \(\theta\) is near zero.

In our exercise, we used: \[\text{lim}_{\theta \to 0} \frac{\theta^n}{\theta^n} = 1 \] for n = 1, 2, 3. These standard limits are fundamental as they simplify complex expressions and allow us to easily deduce the behavior of functions around specific points.

Now, continuity means that a function doesn't have any abrupt jumps or breaks at a point. For example, a function f(x) is continuous at x = a if:

• f(a) is defined
• \(\text{lim}_{x \to a} f(x)\) exists
• \(\text{lim}_{x \to a} f(x) = f(a)\)

For our problem, we rely heavily on the concept of limits as we explore behavior around \(\theta = 0\). These principles help establish the foundation necessary to work through the given exercise.

Taylor Series Expansion

Taylor series expansion is a tool that allows us to approximate complex functions using polynomials. For a function f(x) around x = a, the expansion is given by: \ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots \.

For small values of \(\theta\), the functions \(\cos(\theta)\) and \(\sin(\theta)\) can be approximated as follows:

• \(\cos(\theta) \thickapprox 1 - \frac{\theta^2}{2}\)
• \(\sin(\theta) \thickapprox \theta - \frac{\theta^3}{6}\)

In our exercise, we substituted these approximations into the limit equation to simplify it: \ \lim _{\theta \rightarrow 0}\left(\frac{1+a \cos\theta}{\theta^{2}}-\frac{b \sin \theta}{\theta^{3}}\right) \ becomes \ \frac{1+a(1-\frac{\theta^2}{2})}{\theta^2} - \frac{b(\theta - \frac{\theta^3}{6})}{\theta^3} \ with these Taylor series.

This substitution helps us simplify our expression step-by-step and evaluate the limit effectively.

Trigonometric limits

Trigonometric limits involve functions like sine, cosine, and tangent, which frequently appear in calculus problems. To solve limits involving trigonometric functions near specific points (like 0), we can utilize approximations and standard results.

For instance, knowing that \ \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1 \ is invaluable. Similarly, we often encounter \ \cos(\theta) \thickapprox 1 - \frac{\theta^2}{2} \ for small angles, which allows us to make our calculations much simpler.

In the given problem, these approximations help to break down the limit expression into manageable parts. After the substitution of Taylor expansions:

• \(\frac{1+a(1-\frac{\theta^2}{2})}{\theta^2} - \frac{b(\theta - \frac{\theta^3}{6})}{\theta^3}\)

becomes straightforward to handle. By analyzing each term separately and then taking the limit as \(\theta \to 0\), we can solve for the required variables.

Understanding trigonometric limits is fundamental for working through the IIT JEE calculus problems, making them less intimidating and more approachable.