Problem 20 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{x \rightarrow \infty} \frac{(\log x)^{2}}{x^{n}}, n>0\) is equal to (A) 1 (B) 0 (C) \(-1\) (D) \(\infty\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the expression and the limit

We are given the expression \( \frac{(\log x)^{2}}{x^{n}} \) and asked to find its limit as \( x \to \infty \). Here, \( n \) is a positive number.

Step 2: Analyze the function growth rates

The natural logarithm \( \log x \) grows much slower compared to the polynomial \( x^n \) as \( x \to \infty \). In this case, \( (\log x)^2 \) grows slower than any polynomial \( x^n \) where \( n > 0 \).

Step 3: Apply L'Hopital's Rule if needed

While it might seem like we can use L'Hopital's Rule since both the numerator and the denominator approach infinity, **L'Hopital's Rule is not necessary** here. The denominator grows drastically faster than the numerator as \( x \to \infty \).

Step 4: Conclude the limit

Since the polynomial \( x^n \) in the denominator grows much faster than \( (\log x)^2 \), the overall fraction approaches zero. Hence, \( \lim _{x \rightarrow \infty} \frac{(\log x)^{2}}{x^{n}} = 0 \).

Key Concepts

L'Hopital's Rule

In calculus, L'Hopital's Rule is a useful tool for evaluating indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with such forms, we can differentiate the numerator and the denominator separately to find the limit.

• Use Cases: Apply when direct substitution into the limit gives an indeterminate form.
• Steps: Differentiate the top and bottom functions separately. Retry substitution afterwards.
• Repeat: If the resulting limit is still indeterminate, L'Hopital's Rule can be applied again.

In our exercise, even though the limit of \( \frac{(\log x)^2}{x^n} \) as \( x \to \infty \) suggests indeterminate conditions, L'Hopital's Rule is unnecessary. The polynomial \( x^n \) grows so much faster than \((\log x)^2\) that an intuitive analysis suffices to conclude the limit as zero.

Natural Logarithm

The natural logarithm, often represented as \(\log x\) or more precisely as \(\ln x\), is a logarithmic function with the base \( e \), where \( e \approx 2.71828\).

• Properties: \(\ln(xy) = \ln(x) + \ln(y)\), \(\ln(1) = 0\), \(\ln(e) = 1\).
• Growth: The natural logarithm grows slower than any polynomial. As \( x \to \infty \), \( \log x \to \infty \) very slowly.

In the expression \((\log x)^2\), the growth is primarily determined by the logarithm's nature. Even when squared, \( (\log x)^2 \) still demonstrates slower growth relative to polynomial expressions like \( x^n \) for any positive \( n \). Understanding this inherent growth property allows us to predict the tendency of a limit involving a natural logarithm.

Polynomial Growth Rate

Polynomials, represented as \( x^n \) where \( n \) is a positive integer, exhibit a characteristic growth pattern as \( x \) increases.

• Growth Rate: Polynomials grow faster than linear, logarithmic, and exponential functions with lower exponents.
• Dominance: As \( x \to \infty \), the higher the exponent \( n \), the faster \( x^n \) grows compared to other functions such as \( (\log x)^2 \).

In our context, the polynomial \( x^n \) in the denominator makes the fraction \( \frac{(\log x)^2}{x^n} \) diminish to zero as the denominator's growth significantly outpaces the numerator. This demonstrates how polynomial growth rates can determine the behavior of limits in comparison to other functional forms.