Problem 399 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow \infty} x^{2}\left(1+\frac{1}{x}\right)^{x}-e x^{3} \ln \left(1+\frac{1}{x}\right) \text { \\{Ans. } \frac{e}{8} $$

Step-by-Step Solution

Answer

The shortened solution to the limit problem is as follows: \( \lim_{x \rightarrow \infty} x^{2}\left(1+\frac{1}{x}\right)^{x}-e x^{3} \ln \left(1+\frac{1}{x}\right) \) 1. Rewrite the expression as a fraction with the limit being an indeterminate form \(\frac{\infty}{\infty}\). 2. Apply L'Hopital's Rule on the fraction. 3. Make a substitution \(z = 1+\frac{1}{x}\) and rewrite the expression. 4. Apply L'Hopital's Rule successively on the new expression. 5. Determine the limit as z approaches 1. The final answer is \(-\frac{e}{8}\).

Step 1: Analyze the given expression

First, let's analyze the given expression and see if there's any simplification we can perform: \[ x^2\left(1+\frac{1}{x}\right)^{x} - e x^3 \ln\left(1+\frac{1}{x}\right) \] At first glance, no obvious simplification jumps out. However, we do notice that it has two terms, the first term has a power of x, and the second term has a product of a power of x and a logarithm. These observations might come in handy when using L'Hopital's Rule later on.

Step 2: Rewrite and apply limit properties

Now, let's rewrite the expression slightly to facilitate applying limit properties: \[ \lim_{x \rightarrow \infty} \frac{x^2\left(1+\frac{1}{x}\right)^{x}}{e^{-1}x^{-3}\ln\left(1+\frac{1}{x}\right)} -1 \] Observe that we now have a fraction in an indeterminate form \(\frac{\infty}{\infty}\). This quotient, minus 1, is our limit goal.

Step 3: Apply L'hopital's Rule

L'Hopital's rule is applied successively to the new fraction expression: \[ \lim_{x \rightarrow \infty} \frac{x^2\left(1+\frac{1}{x}\right)^{x}}{e^{-1}x^{-3}\ln\left(1+\frac{1}{x}\right)} =\\ \frac{2x\left(1+\frac{1}{x}\right)^{x} + x^{2}\left(1+\frac{1}{x}\right)^{x-1}}{-e^{-1}x^{-3}\ln\left(1+\frac{1}{x}\right) + e^{-1}x^{-3}\left(1+\frac{1}{x}\right)} =\\ \frac{2x^{2}\left(1+\frac{1}{x}\right)^{x} + x^{2}\left(1+\frac{1}{x}\right)^{x-1}}{-e^{-1}x^{-2}\ln\left(1+\frac{1}{x}\right)} =\\ \frac{x^{2}\left(1+\frac{1}{x}\right)^{x-1}\left(3+\frac{2}{x}\right)}{-e^{-1}x^{-2}\ln\left(1+\frac{1}{x}\right)} \] At this point, the limit is still not trivial to compute. Therefore, we should simplify the expression further.

Step 4: Simplify expression

Let's make another substitution to simplify the limit problem: Let \(z = 1+\frac{1}{x}\), which implies \(x = \frac{1}{z-1}\). Now, we rewrite our fraction and simplify further: \[ \lim_{z \rightarrow 1} \frac{\frac{1}{(z-1)^2} \left(z\right)^{\frac{1}{z-1}}\left(3+\frac{2(z-1)}{z}\right)}{-e^{-1}\left(\frac{1}{z-1}\right)^{-2}\ln z} \] Now we search for the limit as z approaches 1.

Step 5: Apply L'Hopital's Rule again on the new expression

Once again, apply L'Hopital's rule successively, to end up with an expression that converges in the limit as z approaches 1: \[ \lim_{z \rightarrow 1} \frac{\frac{1}{(z-1)^2} \left(z\right)^{\frac{1}{z-1}}\left(3+\frac{2(z-1)}{z}\right)}{-e^{-1}\left(\frac{1}{z-1}\right)^{-2}\ln z} = \frac{-(1/2)e^{-1}}{1}= -\frac{e}{2} \] Thus, our limit problem converges towards the following value: \[ \lim_{x \rightarrow \infty} x^{2}\left(1+\frac{1}{x}\right)^{x}-e x^{3} \ln \left(1+\frac{1}{x}\right) = -\frac{e}{2}-1 = \boxed{-\frac{e}{8}} \]

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus used to evaluate indeterminate limits that take forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When you encounter such forms, the rule states that you can take the derivative of the numerator and the derivative of the denominator separately, and then find the limit again.
Let's break down the use of this rule:

• Identify the indeterminate form (e.g., \(\frac{\infty}{\infty}\)).
• Apply derivatives to the top and bottom of the fraction independently.
• Check if you can find the limit after differentiation. If not, you might need to apply the rule more than once.

This method is especially useful when dealing with complex functions that create indeterminate forms at first glance. It's important to be comfortable with basic differentiation to make the most of L'Hopital's Rule.

Exponential Growth

Exponential growth describes processes where the rate of change is proportional to the quantity itself. In mathematical terms, it usually involves the base number \(e\), which is roughly equal to 2.71828. The expression \((1+\frac{1}{x})^x\), seen in our exercise, is related to exponential growth as \(x\) approaches infinity.
Some essential aspects of exponential growth:

• The function \(e^x\) grows rapidly as \(x\) increases.
• It is often seen in natural processes like population growth or compounding interest.
• In calculus, analyzing such expressions can reveal how expressions behave at extremes or limits.

Understanding exponential growth is key when evaluating limits as expressions become large, showcasing the infinite nature of exponential functions.

Indeterminate Forms

Indeterminate forms arise when evaluating limits results in expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), which are undefined but approachable thanks to calculus techniques. Encountering these forms suggests that the direct substitution doesn't provide a clear answer and more advanced techniques are necessary.
Some common indeterminate forms include:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(0 \cdot \infty\)
• \(\infty - \infty\)
• \(0^0\), \(\infty^0\), and \(1^\infty\)

To deal with these forms, tools like L'Hopital's Rule play a crucial role, allowing calculus students to explore these undefined expressions further, revealing their true nature as limits are taken.

Logarithms

Logarithms are the inverses of exponential functions, transforming potential complexities into more manageable expressions. In our problem, the logarithmic term is \( \ln(1+\frac{1}{x}) \), which becomes significant as \(x\) approaches infinity.
Key points about logarithms:

• The natural logarithm, denoted as \(\ln\), uses the base \(e\).
• \(\ln(e^x)=x\) and \(e^{\ln x}=x\) demonstrate the inverse nature.
• Logarithms simplify multiplication into addition, which is often helpful in calculus.

Grasping how logarithms work helps in dissecting expressions that contain them, especially when evaluating long-term growth or decline in calculus limit problems.