Problem 416 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{e^{\frac{a}{x}}-e^{-\frac{a}{x}}}{e^{\frac{a}{x}}+e^{-\frac{a}{x}}}(a>0) .\\{\text { Ans. } 1,-1\\} $$

Step-by-Step Solution

Answer

The short answer is: \(\lim_{x \rightarrow 0} \frac{e^{\frac{a}{x}} - e^{-\frac{a}{x}}}{e^{\frac{a}{x}} + e^{-\frac{a}{x}}} = 1\).

Step 1: Recognizing Hyperbolic Sine and Cosine Functions

Recall that the hyperbolic sine and cosine functions, usually denoted as \(sinh\) and \(cosh\), can be defined as follows: \[ \sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2}. \] We rewrite the expression in terms of sinh and cosh: \[ \frac{\frac{e^{\frac{a}{x}} - e^{-\frac{a}{x}}}{2}}{\frac{e^{\frac{a}{x}} + e^{-\frac{a}{x}}}{2}} = \frac{\sinh \left(\frac{a}{x}\right)}{\cosh \left(\frac{a}{x}\right)}. \] Now the expression has been simplified, and we proceed to finding the limit.

Step 2: Evaluating the Limit

We want to find the limit of the function as \(x\) approaches 0: \[ \lim_{x \rightarrow 0} \frac{\sinh \left(\frac{a}{x}\right)}{\cosh \left(\frac{a}{x}\right)}. \] Notice that this is still an indeterminate form because as \(x\) approaches 0, \(\frac{a}{x}\) approaches infinity. We will apply L'Hôpital's Rule to evaluate the limit.

Step 3: Applying L'Hôpital's Rule

As \(x\) approaches 0, \(\frac{a}{x}\) approaches infinity. So we can rewrite the limit as: \[ \lim_{y \rightarrow \infty} \frac{\sinh(y)}{\cosh(y)}, \] where \(y = \frac{a}{x}\). Now, we apply L'Hôpital's Rule, which gives us: \[ \lim_{y \rightarrow \infty} \frac{\cosh(y)}{\sinh(y)} =\lim_{y\rightarrow \infty} \frac{1}{\tanh(y)}, \] since \(\frac{d(\sinh y)}{dy} = \cosh y\) and \(\frac{d(\cosh y)}{dy} = \sinh y\). Recall that: \[ \tanh(y) = \frac{\sinh(y)}{\cosh(y)}, \] and as \(y\) approaches infinity, the denominator (\(\cosh y\)) grows much faster than the numerator (\(\sinh y\)). Hence, \(\tanh(y)\) approaches 1 as \(y\) approaches infinity.

Step 4: Final Answer

Since the limit of \(\frac{1}{\tanh(y)}\) as \(y\) approaches infinity is 1, \[ \lim_{x \rightarrow 0} \frac{e^{\frac{a}{x}} - e^{-\frac{a}{x}}}{e^{\frac{a}{x}} + e^{-\frac{a}{x}}} = 1. \] Thus, the final answer is 1.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are significant in mathematics, often found in calculus. These functions resemble trigonometric functions but are based on hyperbolas rather than circles. The two primary hyperbolic functions are the hyperbolic sine, \( ext{sinh}\), and hyperbolic cosine, \( ext{cosh}\). These are defined as:

• \( ext{sinh} \, x = \frac{e^x - e^{-x}}{2}\)
• \( ext{cosh} \, x = \frac{e^x + e^{-x}}{2}\)

Hyperbolic functions have many applications, ranging from geometry to physics, and are essential for solving certain types of equations. In calculus, they sometimes appear when dealing with limits or integrals involving exponential functions.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to evaluate limits that result in indeterminate forms, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule states that for functions \(f(x)\) and \(g(x)\), the limit \[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} \] can be found by differentiating the numerator and the denominator separately: \[ \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}, \] provided this new limit exists.
Applying L'Hôpital's Rule repeatedly may sometimes be needed if the first derivative evaluation still results in an indeterminate form.

Indeterminate Forms

Indeterminate forms are expressions that arise in calculus when evaluating limits, which do not have a defined answer directly. Common indeterminate forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(0^0\), \(1^\infty\), and \(\infty^0\). When these forms appear, additional analysis, such as factoring, conjugates, or L'Hôpital's Rule, may be necessary. The problem from the exercise above originally presents an indeterminate form, \(\frac{\infty}{\infty}\), when substituting values directly, signaling the need for further evaluation using techniques like L'Hôpital's Rule.

Tanh Function

The hyperbolic tangent function, \(\tanh\), is a hyperbolic function derived from hyperbolic sine and cosine. It is defined as: \[ \tanh \, x = \frac{\sinh \, x}{\cosh \, x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \] \(\tanh(x)\) is useful for describing growth processes and other phenomena in mathematics and physics.
In the context of the given exercise, as \(x\) approaches zero, and hence \(y = \frac{a}{x}\) approaches infinity, the function \(\frac{\sinh(y)}{\cosh(y)}\) simplifies to \(\tanh(y)\), which approaches 1. This simplification was essential in evaluating the limit as exemplified in the solution.`