Problem 301 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}}\left\\{\text { Ans. } e^{a b}\right\\} $$

Step-by-Step Solution

Answer

The short answer for the given limit problem is the value of the limit, which is: \(\lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}} = e^{ab}\).

Step 1: 1. Rewrite the expression in exponential form

The given expression is an exponential function. We can rewrite the limit expression in exponential form for simplification: \(\lim _{x \rightarrow 0}e^{\frac{1}{x}\ln(\cos x+a \sin b x)}\)

Step 2: 2. Apply L'Hopital's Rule

We now need to find the derivative of the exponent, \(\frac{1}{x}\ln(\cos x+a \sin b x)\), with respect to x. L'Hopital's rule requires having the form \(\frac{f(x)}{g(x)}\). To apply L'Hopital's Rule, let's rewrite the expression as follows: \(\lim_{x \rightarrow 0} e^{\frac{\ln(\cos x+a \sin b x)}{x}}\) Now, applying L'Hopital's Rule, we need to find the derivative of numerator and the derivative of denominator with respect to x: Numerator derivative: \(\frac{d}{dx} [\ln(\cos x+a \sin b x)] = \frac{-\sin x + ab \cos bx}{\cos x+a \sin b x}\) Denominator derivative: \(\frac{d}{dx} [x] = 1\)

Step 3: 3. Compute the limit

Now that we have the derivatives of the numerator and denominator, let's compute the limit: \(\lim_{x \rightarrow 0} e^{\frac{\frac{-\sin x + ab \cos bx}{\cos x+a \sin b x}}{1}} = e^{\frac{-\sin x + ab \cos bx}{\cos x+a \sin b x}}\) As \(x \rightarrow 0\), the expression simplifies to: \(e^{\frac{-(0) + ab \cos b(0)}{\cos (0)+a \sin b(0)}} = e^{\frac{ab \cdot 1}{1 + a \cdot 0}} = e^{ab}\) Therefore, the value of the limit is: \(\lim _{x \rightarrow 0}(\cos x+a \sin b x)^{\frac{1}{x}} = e^{ab}\)

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus used to determine limits that have indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When you encounter a limit with these indeterminate forms, you can apply L'Hopital's Rule to simplify the calculation. Let's break down how it works:

• If \(\lim_{x \to c} f(x) = 0\) and \(\lim_{x \to c} g(x) = 0\), or both approach infinity, then:

\[\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\]

• Here, \(f'(x)\) and \(g'(x)\) are the derivatives of \(f(x)\) and \(g(x)\) respectively.
• If the original limit still results in an indeterminate form, apply L'Hopital's Rule again.
• L'Hopital's Rule works when both functions involved are differentiable around the point of interest.

In our exercise, we applied L'Hopital's Rule by turning the limit into a fraction suitable for differentiation. It helped us find the limit where simpler algebraic approaches would have been cumbersome.

Exponential Functions

Exponential functions are fundamental in calculus and mathematics. They have a form like \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718. These functions grow rapidly and appear in real-world applications like population growth, radioactive decay, and interest calculations.

When dealing with limits involving exponential functions, especially those in the form \(e^{f(x)}\), it’s helpful to use logarithms to simplify the expression. In our problem, the expression \((\cos x + a \sin bx)^{\frac{1}{x}}\) was rewritten as an exponential function, making use of the identity:

• \((...)= e^{\ln(...)},\)
• which allows us to transform the expression into \(e^{\frac{1}{x} \ln(\cos x + a \sin bx)}\).

This technique is useful because exponential functions are smooth and differentiable, making them easier to handle algebraically. They allow us to apply rules like L'Hopital's more effectively and reach a solution faster.

Trigonometric Functions

Trigonometric functions such as sine and cosine are paramount in calculus. They describe periodic phenomena and are crucial in a variety of applications ranging from physics to engineering.

In our problem, we use the trigonometric functions \(\cos x\) and \(\sin bx\) inside a limit. These functions help illustrate:

• The interplay between oscillation (sine function) and rotation (cosine function).
• The importance of having a handle on their derivatives: \(\cos x\) differentiates to \(-\sin x\), and \(\sin bx\) differentiates to \(b \cos bx\).

Understanding these derivatives is critical when applying L'Hopital's Rule, as seen in the exercise where we computed the derivative of the logarithmic part inside the limit.These functions are both periodic, so they behave predictably as \(x \to 0\). This consistency enables us to evaluate limits more precisely, providing a clear path to the solution \(e^{ab}\).