Problem 69 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The jump of the function at the point of discontinuity i.e., \(x=1\) of the function \(f(x)=\lim _{n \rightarrow \infty} \frac{\log (2+x)-x^{2 n} \sin x}{1+x^{2 n}}\) is (A) \(\sin 1-\log 3\) (B) \(\sin 1+\log 3\) (C) \(-\sin 1+\log 3\) (D) None of these

Step-by-Step Solution

Answer

The jump at the point of discontinuity is \(-\sin 1 - \log 3\). Option (A) is correct.

Step 1: Understand the Problem

We need to find the jump of the function at the point of discontinuity, specifically at \(x = 1\). The function is given by a limit as \(n\) approaches infinity: \[ f(x) = \lim_{n \rightarrow \infty} \frac{\log(2+x) - x^{2n} \sin x}{1 + x^{2n}}. \] We will evaluate the left-hand limit and the right-hand limit at \(x = 1\) and find their difference.

Step 2: Evaluate Left-Hand Limit

Consider \(x = 1^-\), which implies values just less than 1. For small negative \(\epsilon\), \[ x = 1 - \epsilon. \]As \(\epsilon \to 0\), \( (1-\epsilon)^{2n} \to 0 \), since \( (1-\epsilon) < 1 \) and powers of numbers less than 1 shrink to zero. Thus, the numerator becomes \( \log(3) - 0 \sin(1-\epsilon) = \log(3) \) and the denominator is \( 1 + 0 = 1 \). Therefore, \[ \lim_{x \to 1^-} f(x) = \frac{\log(3)}{1} = \log(3). \]

Step 3: Evaluate Right-Hand Limit

Now consider \(x = 1^+\), which implies values just greater than 1. For small positive \(\epsilon\), \[ x = 1 + \epsilon. \]As \(\epsilon \to 0\), \( (1+\epsilon)^{2n} \to \infty \), since \( (1+\epsilon) > 1 \) and powers of numbers greater than 1 grow to infinity. The term \(x^{2n} \to \infty\) dominates so the function behaves like \[-\sin(1+\epsilon), \] because \(-x^{2n} \sin(1+\epsilon)\) becomes the dominant term. The denominator approximates to \(\infty\) too. Thus, \[ \lim_{x \to 1^+} f(x) = \lim \left(-\sin(1+\epsilon)\right) = -\sin 1. \]

Step 4: Calculate Jump at Discontinuity

The jump of the function at \(x = 1\) is calculated as the difference between the right-hand limit and left-hand limit:\[ \text{Jump} = \lim_{x \to 1^+} f(x) - \lim_{x \to 1^-} f(x) = -\sin(1) - \log(3). \] Thus, the jump is \[ -\sin 1 - \log 3. \]

Step 5: Choose the Correct Option

The final answer corresponds to the calculated jump. The options provided are:- (A) \(-\sin 1 - \log 3\)Since this matches with our calculated jump \(-\sin 1 - \log 3\), the correct option is (A).

Key Concepts

Limits

The concept of a limit is crucial in calculus. It helps us understand the behavior of a function as it approaches a certain point. When dealing with functions, especially those involving infinities, limits provide a precise method for evaluating behavior close to a certain point without necessarily reaching it.

To visualize, consider a function that gets closer and closer to a numerical value as its input approaches a point. This value is the limit. Limits are foundational for defining derivatives and integrals too.

Understanding limits involves looking at both sides of a point to see if the function approaches the same value. If so, the limit exists and it is that value. In exercises involving discontinuity, limits can tell us how the function behaves just before and just after a certain point.

Left-Hand Limit

When we talk about the left-hand limit of a function at a point, we're interested in the function's behavior as it approaches the point from the left side. Imagine walking towards a point on the graph from negative values of the x-axis. This is the left-hand side.

Mathematically, if you want to evaluate the left-hand limit at a point like 1, you consider points that are just slightly less than 1, often noted as approaching 1 from the left. An example notation is:

• \( \lim_{x \to 1^-} f(x) \)

This expression specifically looks at what happens when values just below the point get ever closer to it, without crossing over. In the given solution, the left-hand limit at \(x = 1\) was found to be \(\log(3)\) because the values below 1 eventually simplify to that result as \(n\) grows.

Right-Hand Limit

Similar to the left-hand limit, the right-hand limit examines the behavior of a function as it approaches a specific point, but this time from the right, or from values that are slightly greater than the point.

To find the right-hand limit of the function as \(x\) approaches 1, one would look at points just above 1. Written in limit notation, this is:

• \( \lim_{x \to 1^+} f(x) \)

Approaching from the right can sometimes show a different character of the function. In the problem at hand, moving towards \(x = 1\) from values slightly larger gave us a limit of \(-\sin(1)\). This shows that the expression drastically changes its behavior around \(x = 1\), as compared to from the left side.

Jump Discontinuity

A jump discontinuity occurs when the left and right-hand limits at a particular point exist but are not equal. This causes the graph to "jump" from one point to another, showing a break or an abrupt change in the value of the function at the point.

If you imagine drawing a continuous line graph, you would have to lift your pen to continue drawing after the jump, indicating a sudden change.

In the given exercise, the left-hand limit is \(\log(3)\) and the right-hand limit is \(-\sin(1)\). The difference between these two results in the 'jump', calculated as the difference:

• Jump = \(\lim_{x \to 1^+} f(x) - \lim_{x \to 1^-} f(x)\)

Thus, a jump of \(-\sin(1) - \log(3)\) occurs, indicating a clear discontinuity present at \(x = 1\), an example of a jump discontinuity.