Problem 398 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \begin{aligned} &\lim _{x \rightarrow 2} \frac{(\sin \theta)^{x}+(\cos \theta)^{x}-1}{x-2} \\ &\text { \\{Ans. } \left.\sin ^{2} \theta \ln (\sin \theta)+\cos ^{2} \theta \ln (\cos \theta)\right\\} \end{aligned} $$

Step-by-Step Solution

Answer

The short answer to the given limit problem is: \(\lim_{x\rightarrow2} \frac{(\sin \theta)^{x}+(\cos \theta)^{x}-1}{x-2} = (\sin \theta)^{2} \ln(\sin \theta)+(\cos \theta)^{2} \ln(\cos \theta)\).

Step 1: Recognize the indeterminate form

Before we start, let's ensure that the limit is of the indeterminate form. As x approaches 2, the numerator approaches: \((\sin \theta)^{2}+(\cos \theta)^{2}-1\), and since \((\sin \theta)^{2}+(\cos \theta)^{2}=1\), the numerator approaches 0. The denominator also approaches 0, as \(x-2\) is 0 when x is 2. Hence, the limit is of the form 0/0, which is indeterminate.

Step 2: L'Hopital's Rule

To apply L'Hopital's rule, we need to find the derivatives of both the numerator and the denominator with respect to x. Let's find the derivatives: Numerator: \(\frac{d}{dx} \left[ (\sin \theta)^{x}+(\cos \theta)^{x}-1 \right]= (\sin \theta)^{x} \ln(\sin \theta)+(\cos \theta)^{x} \ln(\cos \theta)\), and, Denominator: \(\frac{d}{dx}(x-2)=1\).

Step 3: Apply L'Hopital's Rule

Now that we have computed the derivatives, let's use L'Hopital's rule to find the limit: \(\lim_{x\rightarrow2} \frac{(\sin \theta)^{x}+(\cos \theta)^{x}-1}{x-2}\) = \(\lim_{x\rightarrow2} \frac{(\sin \theta)^{x} \ln(\sin \theta)+(\cos \theta)^{x} \ln(\cos \theta)}{1}\) Since the denominator is 1, the limit simplifies to: \(\lim_{x\rightarrow2} [(\sin \theta)^{x} \ln(\sin \theta)+(\cos \theta)^{x} \ln(\cos \theta)]\)

Step 4: Substitute x with 2

Now, substitute x with 2 in the simplified limit: \(\lim_{x\rightarrow2} [(\sin \theta)^{x} \ln(\sin \theta)+(\cos \theta)^{x} \ln(\cos \theta)]\) = \((\sin \theta)^{2} \ln(\sin \theta)+(\cos \theta)^{2} \ln(\cos \theta)\) Hence, the limit of the given function as x approaches 2 is \((\sin \theta)^{2} \ln(\sin \theta)+(\cos \theta)^{2} \ln(\cos \theta)\).

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form occurs when you directly substitute a value into a mathematical expression and you end up with an ambiguous condition. These forms usually appear similar to "0/0", "∞/∞", "0\cdot∞", and others. The form "0/0" is perhaps the most common and is called an indeterminate form because it doesn't provide a clear answer or limit.

• An indeterminate form means that the limit cannot be simply found by direct substitution.
• You need special techniques, such as L'Hopital’s Rule, to resolve such indeterminate forms.

In the exercise above, as \(x\) approaches \(2\), both the numerator and the denominator tend to zero. This results in the form "0/0".

This is characteristic of an indeterminate form, which indicates that we cannot immediately find the limit without further manipulation or analysis, precisely where L'Hopital’s Rule comes into play.

Limits in Calculus

Limits in calculus help us understand the behavior of functions as they approach specific points or infinity. They form a foundational concept in calculus, underlying much of the theory and application in areas like derivatives and integrals.

• Finding limits is essential because they allow mathematicians to define the notion of continuity, derivatives, and integrals.
• When dealing with expressions that become indeterminate, limits provide a systematic way to evaluate these expressions effectively.
• L'Hopital's Rule is an important tool used when direct substitution in limit expressions results in an indeterminate form.

In the solution, approaching the problem with limits enables the determination of the expression’s value as \(x\) nears \(2\). This approach is necessary given the indeterminate form present in the expression. Without the concept of limits, evaluating such complex situations would be extremely challenging.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are crucial in mathematics. They describe the relationships between angles and sides of triangles and are also periodic functions, making them applicable in various periodic scenarios like waves.

• These functions are vital in calculus, particularly when dealing with oscillatory functions and periodic behavior.
• In calculus, it's crucial to understand the properties of trigonometric functions, such as their limits and derivatives.

In the exercise provided, the trigonometric functions sin and cos are raised to the power of \(x\). Their properties are used effectively in calculating the limit with L'Hopital's Rule. The expressions \((\sin \theta)^x\) and \((\cos \theta)^x\) are fundamental to the problem's solution, and understanding them helps in finding the limit.