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If \(\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=2\), then \(\lim _{x \rightarrow 0}[f(x)]\) equals, (where [.] denotes the greatest integer function) (1) 2 (2) 1 (3) 0 (4) limit does not exist

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Step by step solution

01

- Understand the given limit

We are given that \(\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=2\). This limit tells us how \(\frac{f(x)}{x^2}\) behaves as \(x\) approaches 0.
02

- Express the limit in terms of \(f(x)\)

From the given limit, \(\lim _{x \rightarrow 0} \frac{f(x)}{x^2} = 2\), we can express it as: \(\lim_{x \rightarrow 0} f(x) = 2x^2\).
03

- Evaluate the limit of \(f(x)\) as \(x\) approaches 0

Substitute \(x\) with 0 in \(2x^2\): \[ \lim_{x \rightarrow 0} f(x) = 2 \times (0^2) = 2 \times 0 = 0 \]
04

- Apply the Greatest Integer Function

The greatest integer function \([.]\) returns the greatest integer less than or equal to a given number. Since \(\lim_{x \rightarrow 0} f(x) = 0\), the greatest integer less than or equal to 0 is 0. Therefore, \([\lim_{x \rightarrow 0} f(x)] = 0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

IIT JEE Calculus Problem
The IIT JEE examination often includes questions on calculus that require critical thinking and a good grasp of fundamental concepts. This particular problem combines the concepts of limit evaluation and the greatest integer function.
When solving such problems:
  • First clearly identify the limit you need to evaluate.
  • Express the given function in a simpler form to make limit evaluation straightforward.
  • Understand the role of any special functions, like the greatest integer function in this case.

By following these steps, you can systematically approach and solve complex calculus problems that are commonly part of the IIT JEE examination.

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Most popular questions from this chapter

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\(\lim _{x \rightarrow \infty} \frac{2 \cdot x^{1 / 2}+3 \cdot x^{1 / 3}+4 \cdot x^{1 / 4}+\ldots . .+n \cdot x^{1 / n}}{(3 x-4)^{1 / 2}+(3 x-4)^{1 / 3}+\ldots+(3 x-4)^{1 / n}}(\) here \(n \in N, n \geq 2)\) is equal to (1) \(\frac{2}{\sqrt{3}}\) (2) \(\frac{\sqrt{3}}{2}\) (3) \(\frac{1}{2}\) (4) \(\frac{1}{\sqrt{3}}\)

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