Problem 278 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\lim _{x \rightarrow \pm \infty}\left(1+\frac{1}{x}\right)^{x^{2}} \text { \\{ Ans. }+\infty, 0\right\\} $$

Step-by-Step Solution

Answer

The short answer to the given limit problem is: As \(x\) approaches positive infinity, the limit is \(+\infty\), and as \(x\) approaches negative infinity, the limit is \(0\).

Step 1: Analyze the limit as x approaches positive infinity

Let's consider the limit as x approaches positive infinity: \(\lim_{x \rightarrow +\infty}\left(1+\frac{1}{x}\right)^{x^{2}}\) **Step 2: Apply L'Hôpital's Rule**

Step 2: Apply L'Hôpital's Rule and simplify the exponent

First, rewrite the expression in a more manageable form so that L'Hôpital's Rule can be used: \(\lim_{x \rightarrow +\infty} \exp\left(x^2 \cdot \ln{\left(1 + \frac{1}{x}\right)}\right)\) Applying L'Hôpital's Rule, we differentiate both exponent and logarithm with respect to \(x\): \(\lim_{x \rightarrow +\infty} \exp\left(x^2 \cdot \frac{1}{1 + \frac{1}{x}} \cdot \frac{-1}{x^2}\right)\) Simplify the expression: \(\lim_{x \rightarrow +\infty} \exp(-1)\) This returns a constant value, so the limit as \(x\) approaches positive infinity is: \(+\infty\) **Step 3: Negative infinity**

Step 3: Analyze the limit as x approaches negative infinity

Next, let's consider the limit as x approaches negative infinity: \(\lim_{x \rightarrow -\infty}\left(1+\frac{1}{x}\right)^{x^{2}}\) **Step 4: Apply L'Hôpital's Rule**

Step 4: Apply L'Hôpital's Rule and simplify the exponent

Rewrite the expression in a more manageable form so that L'Hôpital's Rule can be used: \(\lim_{x \rightarrow -\infty} \exp\left(x^2 \cdot \ln{\left(1 + \frac{1}{x}\right)}\right)\) Again, applying L'Hôpital's Rule, we differentiate both exponent and logarithm with respect to \(x\): \(\lim_{x \rightarrow -\infty} \exp\left(x^2 \cdot \frac{1}{1 + \frac{1}{x}} \cdot \frac{-1}{x^2}\right)\) Simplify the expression: \(\lim_{x \rightarrow -\infty} \exp(1)\) This returns a constant value, so the limit as \(x\) approaches negative infinity is: \(0\) So, the final answer is: \(+\infty, 0\)

Key Concepts

Infinite Limits

Infinite limits refer to the behavior of a function as the input grows very large or very small, essentially moving towards positive or negative infinity. In calculus, determining the behavior of functions as they approach infinity is key to understanding their nature at extreme values. For example, when we have a limit like \( \lim_{x \rightarrow \pm \infty} f(x) \), it describes how the function \( f(x) \) behaves as \( x \) either becomes very large positively or negatively.
When computing these limits, the result might indicate that the function grows indefinitely, leading to an infinite limit, or it might approach a finite number or zero. This concept helps identify the asymptotic behavior of functions and is critical in fields such as physics and engineering where predicting long-term behavior is essential.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits that initially present as indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule provides a systematic way to resolve these indeterminate forms by differentiating the numerator and the denominator, ideally simplifying the expression to a more solvable form.
To apply L'Hôpital's Rule, one must ensure both the numerator and denominator approach zero or infinity. If that condition is met, the rule can be applied repeatedly if necessary until a determinate form is reached. In our exercise, we encounter such a scenario with limits involving exponentials. By transforming the expression using natural logarithms and the exponential function, L'Hôpital's Rule is applied to tackle the complex relationship within the exponent. This method eventually simplifies the expression to reveal a constant limit, illustrating the utility of the rule in simplifying complex limit problems.

Exponential Functions

Exponential functions are a class of mathematical functions characterized by a constant base raised to a variable exponent, commonly shown in the form \( a^x \). They exhibit rapid growth or decay, depending on the base's value, and are defined for all real numbers. The natural exponential function \( e^x \), where \( e \approx 2.71828 \), is particularly significant due to its natural occurrence in various mathematical and real-world phenomena.
Exponential functions are crucial for solving problems involving growth processes like population dynamics, radioactive decay, or financial interests. In the context of our exercise, exponential functions are also used to transform complex expressions into a manageable form, making it easier to calculate limits. Logarithms often accompany exponentials during such transformations, allowing us to leverage properties that simplify the computational process through methods like L'Hôpital's Rule.