Problem 27 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

\(\lim _{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right)\) is equal to (a) 2 (b) \(1 / 2\) (c) \(\infty\) (d) 0

Step-by-Step Solution

Answer

(a) 2

Step 1: Identify the Limit Expression

The given expression is \( \lim _{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right) \). Our task is to determine this limit as \( x \) approaches infinity.

Step 2: Simplify the Expression

Notice that as \( x \rightarrow \infty \), the expression \( \frac{2}{x} \rightarrow 0\). We can simplify the expression by considering a substitution where \( y = \frac{2}{x} \). Then as \( x \rightarrow \infty \), \( y \rightarrow 0^+ \). This transforms our limit into \( \lim_{y \rightarrow 0^+} \frac{2}{y} \sin(y) \).

Step 3: L'Hopital's Rule

The expression \( \lim_{y \rightarrow 0} \frac{2 \sin(y)}{y} \) can be rewritten in the form \( \frac{0}{0} \), an indeterminate form suitable for L'Hopital's Rule. L'Hopital's Rule allows us to differentiate the numerator and the denominator. Differentiate to get \( \frac{2\cos(y)}{1} \).

Step 4: Evaluate the Limit using L'Hopital's Rule

Evaluate the limit of the new expression: \( \lim_{y \rightarrow 0} 2\cos(y) \). As \( y \rightarrow 0 \), \( \cos(y) \rightarrow \cos(0) = 1 \). Hence, \( \lim_{y \rightarrow 0} 2\cos(y) = 2 \).

Step 5: Conclusion

After applying L'Hopital's Rule, we find that \( \lim_{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right) = 2 \). The answer is (a) 2.

Key Concepts

L'Hopital's Rule

When you encounter a limit problem, sometimes the direct substitution leads to an undefined expression. This is where L'Hopital's Rule shines as a powerful tool. Imagine you face an expression that results in forms like \(0/0\) or \(\infty/\infty\) upon substitution. Here, L'Hopital's Rule offers a clever way out.

The rule states that if you have a fraction \(\frac{f(x)}{g(x)}\) and both the numerator and the denominator approach zero or infinity as \(x\) approaches a limit, you can differentiate both the top \(f'(x)\) and bottom \(g'(x)\). Then, take the limit again:

• Only apply L'Hopital's Rule to indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Differentiate the numerator and the denominator separately.
• Evaluate the limit of the new fraction \(\frac{f'(x)}{g'(x)}\).

Sometimes, you might need to apply the rule more than once to resolve the expression into a determinate form.

Indeterminate Forms

Indeterminate forms appear frequently in calculus problems involving limits, and they signal that direct evaluation won't work. For instance, consider the form \(0/0\), which means both the numerator and the denominator are getting very small as \(x\) approaches a certain value. Similarly, forms like \(\infty/\infty\), \(0\cdot\infty\), \(\infty - \infty\), and \(1^\infty\) are other examples.

These forms are like a missing puzzle piece - they need further calculation to find a meaningful result.

To solve these, you need structured techniques like:

• L'Hopital's Rule: Perfect for fractions leading to forms \(0/0\) or \(\infty/\infty\).
• Algebraic manipulation: Restructuring the expression can sometimes turn an indeterminate form into a determinate one.
• Substitution: Making a strategic substitution can simplify and solve the form.

Indeterminate forms are a cue that the straightforward path won't work, inviting you to explore deeper methods for solutions.

Trigonometric Limits

Trigonometric functions can be tricky to work with in limit problems, but understanding their behavior as values approach certain limits is key. One of the most famous limits is the sine limit:

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

This result is foundational and helps tremendously in solving various problems quickly.

When solving trigonometric limits, keep these tips in mind:

• Recognize common limits, like the sine limit, that can simplify expressions dramatically.
• Use identities, such as \(\sin^2(x) + \cos^2(x) = 1\), to help manipulate expressions into more solvable forms.
• Substitution can also break down complex limits into simpler components, making the problem more approachable.
• Sometimes, just rewriting expressions using known limits and identities can lead to quick solutions.

Grasping these strategies can make trigonometric limit calculations far more intuitive and less cumbersome, transforming potentially intimidating problems into straightforward solutions.