Problem 58 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\lim _{x \rightarrow 0}\left(\left[\frac{100 x}{\sin x}\right]+\left[\frac{99 \sin x}{x}\right]\right)\), where [-] represents greatest integer function, is (A) 199 (B) 198 (C) 0 (D) None of these

Step-by-Step Solution

Answer

The value of the limit is 197, which matches option (D) None of these.

Step 1: Simplify the expression

Break down the given limit expression: \[\lim_{x \to 0}\left(\left[\frac{100x}{\sin x}\right] + \left[\frac{99\sin x}{x}\right]\right)\] Using L'Hopital's Rule, since both \ \(\frac{x}{\sin x}\rightarrow 1\ \) and \ \(\frac{\sin x}{x}\rightarrow 1\ \) as \ \(x \rightarrow 0\ \).To simplify, we will explore these limits and solve each part.

Step 2: Evaluate individual limits

Evaluate the expression \[\frac{100x}{\sin x} \approx 100\frac{x}{x}=100\] and, \[\frac{99\sin x}{x} \approx 99\frac{x}{x}=99\] as \(x\rightarrow 0\), both these terms approach their respective integer values.

Step 3: Apply Greatest Integer Function

Taking the floor function of the approximate values:\[\left[\frac{100x}{\sin x}\right] \rightarrow [100] = 99\]and\[\left[\frac{99\sin x}{x}\right] \rightarrow [99] = 98\]when approaching from the lower side.

Step 4: Combine the evaluated expressions

Add the results of the floor functions obtained in Step 3:\[[\frac{100x}{\sin x}] + [\frac{99\sin x}{x}] = 99 + 98 = 197\]

Step 5: Conclusion and verification

Thus, the expression evaluates to 197 when applying the limit calculation for \(x\rightarrow 0\). Therefore, the answer is not 199 or 198.

Key Concepts

Greatest Integer Function

The Greatest Integer Function, often called the floor function, is a mathematical function that takes a real number and gives the greatest integer less than or equal to it. For example:

• For the input 3.7, the output is 3.
• For the input -2.1, the output is -3.
• For the input 5, the output is 5, since 5 is already an integer.

The function notation used is usually \([x]\), where \([x]\) denotes the floor of \(x\). When dealing with limits and the Greatest Integer Function, the crucial point is how close values approach an integer from below. During evaluations, especially involving trigonometric and reciprocal expressions, precision becomes essential to determine which integer the floor function will output. Understanding the behavior around integers helps when calculating limits with floor functions in calculus.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus used to find limits of indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When faced with these forms, L'Hopital's Rule states that:

• If \(\lim_{{x \to a}}f(x) = \lim_{{x \to a}}g(x) = 0\) or \(\pm \infty\), then \[\lim_{{x \to a}}\frac{f(x)}{g(x)} = \lim_{{x \to a}}\frac{f'(x)}{g'(x)}\] provided the limit on the right exists.

This rule turns an indeterminate form into something more manageable by differentiating the numerator and the denominator independently. When applying L'Hopital's Rule, it's crucial to check the conditions and limit form initially. Then the differentiation should lead to an expression where a standard limit evaluation gives a definite answer. Also, it's essential to remember that the rule may need to be applied repeatedly if the resulting limit is still indeterminate after the first application.

Trigonometric Limits

Trigonometric limits are a key concept in calculus, especially those involving the \(\sin(x)\) and \(\cos(x)\) functions. A frequently used result is:

• \(\lim_{{x \to 0}}\frac{\sin x}{x} = 1\)
• \(\lim_{{x \to 0}}\frac{x}{\sin x} = 1\)

These limits form the foundation for many calculus problems involving trigonometric functions. They are derived from the small angle approximation and the squeeze theorem. The fact that both limits equal 1 simplifies many expressions when \(x\) is near zero. When solving problems, knowing these fundamental limits allows us to manage expressions that might otherwise seem complex. By breaking a problem down using these known limits, we can transform a challenging calculus problem into a simpler arithmetic exercise. Understanding these key results will enable students to handle more complex trigonometric limits efficiently.