Problem 78 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{x \rightarrow 0}\left[\frac{a \sin x}{x}\right]+\left[\frac{b \tan x}{x}\right]\), where \(a, b\) are integers and [] denotes integral part, is equal to (A) \(a+b\) (B) \(a+b-1\) (C) \(a-b\) (D) \(a-b-1\)

Step-by-Step Solution

Answer

(A) a+b

Step 1: Understanding the Limits

The problem involves evaluating the limit of two expressions as \(x\) approaches 0, specifically \(\lim _{x \to 0} \left[ \frac{a \sin x}{x} \right] + \left[ \frac{b \tan x}{x} \right] \). The goal is to find the limit of these expressions and examine their integer parts, denoted by the brackets \([ \cdot ]\). This is crucial because the integer part may impact the limit when evaluated.

Step 2: Evaluate Each Limit Component

First, evaluate \(\lim _{x \to 0} \frac{\sin x}{x}\). This is a standard limit that equals 1. Therefore, \(\lim _{x \to 0} \frac{a \sin x}{x} = a\cdot 1 = a\). Next, evaluate \(\lim _{x \to 0} \frac{\tan x}{x}\). This limit also equals 1, resulting in \(\lim _{x \to 0} \frac{b \tan x}{x} = b\cdot 1 = b\).

Step 3: Determine the Integer Parts

The integer part of \(a\) is \([a]\) and for \(b\) is \([b]\). Since the values are approaching \(a\) and \(b\) exactly, as integers, \([a] = a\) and \([b] = b\). This means the integer parts of \(\frac{a \sin x}{x}\) and \(\frac{b \tan x}{x}\) approach \(a\) and \(b\), respectively.

Step 4: Combine Integer Parts

Now, consider \([a]\) and \([b]\). The sum is \([a]+[b] = a + b\). Most importantly, this still holds true since we are dealing directly with the integer parts.

Key Concepts

Trigonometric Limits

Trigonometric limits are a foundation in Calculus, especially when evaluating limits as variables approach zero. Two of the most common trigonometric limits you encounter are \( \lim_{x \to 0} \frac{\sin x}{x} \) and \( \lim_{x \to 0} \frac{\tan x}{x} \). Both of these limits evaluate to 1. This is because as \(x\) approaches zero, both \(\sin x\) and \(\tan x\) gain values nearly identical to \(x\) itself, resulting in fractions approximating \(\frac{x}{x} = 1\).

• Understanding these fundamental trigonometric limits helps us simplify complex limit problems by substituting known results into these common forms.
• Whenever a trigonometric function divided by \(x\) appears in a limit, you can often directly use these limit forms to simplify your calculations.

Applying these standard results in limit problems involving sine or tangent can drastically reduce the complexity and save time.

Integral Part Function

The integral part function, often represented by square brackets \([ \cdot ]\), is used to denote the greatest integer less than or equal to a given number. This function essentially "truncates" a number to its integer component.

• For a positive number, the integral part function rounds down to the nearest whole number.
• For instance, \([3.7] = 3\) and \([-2.3] = -3\).

Using this function in limit evaluation problems can make the step between continuous and discrete mathematics clearer. It influences the final limit by constraining it to whole numbers, as seen in the solution when evaluating \([a]\) and \([b]\) since \(a\) and \(b\) are integers. This may seem trivial when \(a\) and \(b\) are exactly integers, but it can greatly impact results when fractional parts are non-zero.

Limits of Trigonometric Functions

When handling the limits of trigonometric functions, especially as they approach zero, understanding their behavior is crucial. The key result in the original problem relies on standard limits of sine and tangent divided by \(x\). Here's how:

• With \(\sin x\) and \(\tan x\) behaving very similarly to \(x\) as \(x \to 0\), the limits simplify to straightforward forms: \( \frac{\sin x}{x} \to 1 \) and \( \frac{\tan x}{x} \to 1\).
• These simplifications directly influence the main limit problem by allowing constants \(a\) and \(b\) to be factored outside the limits, giving final results directly in terms of these constants.

Recognizing when to employ these standard limits helps solve complex limits effortlessly, enabling you to focus on the other parts of the problem like combining results and understanding impact of the integral part function.