Problem 13 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \([\mathrm{x}]\) denote the greatest integer less than or equal to \(\mathrm{x}\). Then : \(\lim _{x \rightarrow 0} \frac{\tan \left(\pi \sin ^{2} x\right)+(|x|-\sin (x[x]))^{2}}{x^{2}}\) [Jan. 11, 2019(I)] (a) does not exist (b) equals \(\pi\) (c) equals \(\pi+1\) (d) equals 0

Step-by-Step Solution

Answer

The limit equals \(\pi + 1\); option (c).

Step 1: Evaluate the Numerator at Limit

Start by analyzing the components in the numerator of the function:\[ \lim_{x \rightarrow 0} [\tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2] \]As \(x\) approaches 0, \(\sin^2 x \approx x^2\). Therefore, \(\tan(\pi \sin^2 x) \approx \tan(\pi x^2) \). As \(x\) converges to 0, \(\tan(\pi x^2) \approx \pi x^2\) since \(\tan(y) \approx y\) for small \(y\). Thus, this part approximately becomes \(\pi x^2\) near 0.

Step 2: Simplify the Greatest Integer Component

Now, consider the second term: \((|x| - \sin(x[x]))^2\). For small values of \(x\) (approaching zero), \([x] = 0\) since we are considering immediate values around 0.Therefore, \(\sin(x[x]) = 0\) when \([x] = 0\), making this term: \(|x|^2 = x^2\) as \(x\) approaches 0.

Step 3: Combine and Simplify the Numerator

Substitute the approximations into the original numerator:\[ \tan(\pi \sin^2 x) + (|x| - \sin(x[x]))^2 \approx \pi x^2 + x^2 = (\pi+1)x^2\]

Step 4: Plug into the Limit Definition

Now, substitute the simplified numerator back into the full limit expression:\[ \lim_{x \rightarrow 0} \frac{(\pi+1)x^2}{x^2} \]This simplifies to:\[ \lim_{x \rightarrow 0} (\pi + 1) \]

Step 5: Solve the Limit

The limit of a constant is simply the constant itself. Thus:\[ \lim_{x \rightarrow 0} (\pi + 1) = \pi + 1 \]

Key Concepts

Limit Calculations

Limiting calculations is an essential concept in calculus that helps us evaluate the behavior of a function as the input, or variable, approaches a specific value. In this exercise, our goal is to find the limit when \( x \) approaches 0 for the given function. Here’s a brief breakdown of handling limits:

• Analyze the components within the limit. For example, evaluating each piece of the function individually as \( x \) nears the limit point.
• Use approximations for small quantities. For instance, for small \( x \), \( \tan(y) \approx y \) helps simplify calculations.
• Combine and simplify. After evaluating individual components, combine them to see the overall behavior of the function.

Breaking down the problem into smaller parts and approximating small values enables us to simplify and calculate limits efficiently.

Trigonometric Functions

Trigonometric functions play a significant role in understanding periodic phenomena and oscillations. In limit calculations, especially around zero, these functions can be approximated. Here’s how trigonometry ties into this exercise:

• The function contains \( \tan(\pi \sin^2 x) \). For small values of \( x \), it's useful to know that \( \sin^2 x \approx x^2 \) which approximates \( \tan(\pi x^2) \) further to \( \pi x^2 \).
• Using trigonometric identities and approximations helps in simplifying the steps, especially when dealing with limits.

Trigonometric approximations, such as \( \tan(y) \approx y \) for small \( y \), simplify complex expressions and make calculations manageable.

Limits and Continuity

Understanding limits is key to grasping the concept of continuity in a function. A function is continuous at a point if the limit as it approaches from both sides equals the function’s value at that point. Here are some key aspects:

• Continuity involves evaluating whether the function changes in a smooth and predictable manner near a point.
• For the given problem, after simplifying, the expression to find the limit was \( \lim_{x \rightarrow 0} \frac{(\pi+1)x^2}{x^2} \), simplifying to \( \pi + 1 \).
• Understanding that if a function’s limit exists at a given point, it often points to a potential continuity at that point.

Calculating limits and understanding the simplifications provide insights into whether a function is continuous at certain points, aiding in predictions about the function's behavior.