Problem 59 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\lim _{x \rightarrow \infty} \frac{x^{5}}{5^{x}}\) is (A) 1 (B) \(-1\) (C) 0 (D) None of these

Step-by-Step Solution

Answer

The limit is 0 (Option C).

Step 1: Identify the form of the limit

We are asked to find the limit of the function \(f(x) = \frac{x^{5}}{5^x}\) as \(x\) approaches infinity. Observing the structure, as \(x\) grows larger, \(x^{5}\) is a polynomial function and \(5^x\) is an exponential function. Since exponential growth outpaces polynomial growth, it suggests the limit may be 0.

Step 2: Apply logarithmic comparison

To rigorously compare the growth rates, we use logarithms. Consider \(\ln(f(x)) = \ln\left(\frac{x^5}{5^x}\right) = 5\ln(x) - x\ln(5)\). As \(x\) approaches infinity, \(x\ln(5)\) dominates because it grows linearly with \(x\), whereas \(5\ln(x)\) grows much slower. Therefore, \(\ln(f(x))\) approaches \(-\infty\), indicating \(f(x)\) approaches 0.

Step 3: Conclude with L'Hôpital's Rule check

Since \(\lim_{x \rightarrow \infty} \frac{x^5}{5^x}\) is in an indeterminate form \(\frac{\infty}{\infty}\), we can use L'Hôpital's Rule by differentiating the numerator and denominator. After applying L'Hôpital's Rule five times, we differentiate \(x^5\) to \(0\) eventually while differentiating \(5^x\) maintains non-zero value, confirming the limit is indeed 0.

Key Concepts

Polynomial vs Exponential Growth

When analyzing limits like \(\lim_{x \to \infty} \frac{x^5}{5^x}\), it's essential to understand how different functions grow as \(x\) becomes very large. A polynomial such as \(x^5\) has terms that increase through multiplication by \(x\). As \(x\) increases, the degree of the polynomial dictates how strongly it grows. However, regardless of the degree, polynomials cannot keep up with the rapid pace of exponential functions.Exponential functions like \(5^x\) grow by repeatedly multiplying a constant base. This means their increase rate compounds, effectively making them much stronger at larger values of \(x\). In conflicts between polynomial and exponential growth, such as the limit scenario here, exponential functions overwhelmingly win, resulting in a situation where the fraction \(\frac{x^5}{5^x}\) tends towards zero due to the dominance of \(5^x\)'s growth.

• Polynomials grow based on variable multiplication. Example: \(x^5\)
• Exponentials grow based on constant multiplication. Example: \(5^x\)
• Exponential growth surpasses polynomial growth as \(x\) increases, impacting limit outcomes significantly.

Logarithmic Comparison

Logarithmic comparison is a helpful tool when analyzing how functions behave, particularly for determining which function grows faster. For our function \(f(x) = \frac{x^5}{5^x}\), a logarithmic transformation can simplify the comparison by converting multiplicative relationships into additive ones.Taking the natural logarithm, \( \ln(f(x)) = \ln\left(\frac{x^5}{5^x}\right) = 5\ln(x) - x\ln(5) \), we can clearly see how the growth rates compare. The term \(x\ln(5)\) increases linearly like \(x\), while \(5\ln(x)\) tweaks more slowly, making it evident that \(x\ln(5)\) will eventually overcome the slower-growing \(5\ln(x)\).

• Logarithms change growth comparisons from multiplicative to additive.
• Comparing growth rates becomes clearer through simpler arithmetic terms.
• In this function, \(x\ln(5)\) quickly overtakes \(5\ln(x)\).

As the logarithm of \(f(x)\) approaches \(-\infty\), \(f(x)\) itself must approach zero, which supports our conclusion about the limit.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool for evaluating limits of ratios that yield indeterminate forms, like \(\frac{\infty}{\infty}\). In our case, the function \(\lim_{x \to \infty} \frac{x^5}{5^x}\) presents such a form.L'Hôpital's Rule tells us that we can differentiate the numerator and the denominator separately to derive a new, simpler form for evaluating the limit. By applying the rule repeatedly, we continue differentiating both parts until we no longer have an indeterminate form.For \(\frac{x^5}{5^x}\):

• First, differentiate the numerator: \(x^5\) becomes \(5x^4\), then \(20x^3\), and so on until it becomes \(0\).
• Meanwhile, the denominator \(5^x\) keeps growing after each differentiation, confirming it's never \(0\).

Through successive differentiation, the numerator diminishes quickly while the denominator remains an exponentially growing entity. This process confirms that the original problematic expression simplifies with L'Hôpital's Rule, converging to a limit of 0.