Problem 373 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

The short answer to the question is: The limit of the function \((e^{x}+x)^{\frac{1}{x}}\) as \(x\) approaches 0 is \(e^2\).

Step 1: Rewrite as a quotient

First, we rewrite the expression as a limit of a quotient: \(\lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\frac{1}{x}}=e^{\lim_{x \rightarrow 0} \frac{1}{x} \ln\left(e^{x}+x\right)}\) #Step 2: Apply L'Hôpital's Rule#

Step 2: Apply L'Hôpital's Rule

Now we need to find the limit of the exponent, which is \(\lim_{x \rightarrow 0} \frac{1}{x} \ln\left(e^{x}+x\right)\). Since we have a quotient, we can apply L'Hôpital's Rule. First, we need to find the derivative of the numerator and denominator: \(\mathrm{numerator} = ln(e^{x}+x)\) \(\mathrm{Denominator} = x\) Taking their derivatives, we get: \(\frac{d(\mathrm{numerator})}{dx} = \frac{e^{x}+1}{e^{x}+x}\) \(\frac{d(\mathrm{Denominator})}{dx} = 1\) Applying L'Hôpital's Rule, we have: \(\lim_{x \rightarrow 0} \frac{1}{x} \ln\left(e^{x}+x\right) = \lim_{x \rightarrow 0} \frac{\frac{e^{x}+1}{e^{x}+x}}{1}\) #Step 3: Evaluate the limit of the quotient#

Step 3: Evaluate the limit of the quotient

Now that we've applied L'Hôpital's Rule, we can evaluate the limit as \(x\) approaches 0: \(\lim_{x \rightarrow 0} \frac{e^{x}+1}{e^{x}+x} = \frac{e^{0}+1}{e^{0}+0} = \frac{2}{1} = 2\) #Step 4: Compute the final answer#

Step 4: Compute the final answer

Now we have the limit of the exponent, so we can find the final answer: \(\lim _{x \rightarrow 0}\left(e^{x}+x\right)^{\frac{1}{x}} = e^{\lim_{x \rightarrow 0} \frac{1}{x} \ln\left(e^{x}+x\right)} = e^2\) #Solution#: The limit of the function \((e^{x}+x)^{\frac{1}{x}}\) as \(x\) approaches 0 is \(e^2\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is an invaluable tool when dealing with limits involving indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). This rule allows you to find limits by taking the derivatives of the numerator and the denominator.

• First, ensure that the limit takes an indeterminate form.
• Take the derivative of the numerator and the denominator.
• Calculate the limit of the new fraction. If the result is still indeterminate, you can apply L'Hôpital's Rule again.

In this particular exercise, the original expression \(\lim_{x \rightarrow 0} \frac{1}{x} \ln(e^{x}+x)\) forms a \(\frac{0}{0}\) indeterminate case. Hence, we differentiate the numerator, \(\ln(e^{x}+x)\), and the denominator, \(x\). The derivatives are \(\frac{e^{x}+1}{e^{x}+x}\) and 1, respectively. Reapplying the limit to this new fraction lets us simplify and solve easily.

Exponential Functions

Exponential functions are a critical concept in mathematics, usually expressed in the form \(y = a^{x}\). In this context, the base \(e\) is often used, where \(e\) is an irrational number approximately equal to 2.718.

• These functions exhibit continual growth or decay.
• The rate of growth is proportional to the value of the function itself.
• Exponential growth occurs when the base is greater than 1, while decay happens if the base is between 0 and 1.

The expression \((e^{x}+x)^{\frac{1}{x}}\) incorporates an exponential component since \(e^{x}\) is part of the inner expression of the power. Here, we utilized properties of exponential functions to rewrite the expression in terms of natural logarithms, paving the way for the application of calculus concepts such as L'Hôpital's Rule.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is the inverse of the exponential function \(e^x\). Understanding natural logarithms is crucial since they often appear in calculus problems.

• Natural logarithms can help solve exponential equations.
• They simplify differentiation and integration of exponential functions.
• The derivative of \(\ln(x)\) is \(\frac{1}{x}\), which helps in solving limits.

In the exercise, \(\ln(e^{x}+x)\) is the numerator function we differentiate when applying L'Hôpital's Rule. This transformation was key since it allowed us to express the original limit problem in a form amenable to standard calculus techniques, ultimately aiding in finding the solution to the limit.