Problem 29 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Discuss the limit, continuity and differentiability of the function $$ \begin{aligned} f(x) &=x\left(\frac{3 e^{\frac{1}{x}}+4}{2-e^{\frac{1}{x}}}\right), x \neq 0 \\\ &=0, \quad x=0 \end{aligned} $$ $$ \text { at } x=0 \text { . } $$

Step-by-Step Solution

Answer

The function \(f(x)\) has a limit of 7 as x approaches 0. However, it is not continuous or differentiable at \(x=0\) as the value of the function at \(x=0\) is not equal to the limit, and the function must be continuous in order to be differentiable.

Step 1: Find the limit as x approaches 0

To find the limit as x approaches 0, we will use L'Hôpital's rule, since we have an indeterminate form of the type $$\frac{0}{0}$$. First, let's rewrite the function as a fraction: $$ g(x) = \frac{3xe^{\frac{1}{x}}+4x}{2x - xe^{\frac{1}{x}}} $$ Now, we will find the derivatives of the numerator and the denominator with respect to x: $$ \begin{aligned} g'(x) &= \frac { \frac{d}{dx}(3xe^{\frac{1}{x}}+4x) } { \frac{d}{dx}(2x - xe^{\frac{1}{x}}) } \\\ \end{aligned} $$ Calculating the derivatives, we have: $$ \begin{aligned} \frac{d}{dx}(3xe^{\frac{1}{x}}) &= 3e^{\frac{1}{x}} - 3xe^{-\frac{1}{x}}\left(-\frac{1}{x^2}\right) \\ \frac{d}{dx}(4x) &= 4 \\ \frac{d}{dx}(2x) &= 2 \\ \frac{d}{dx}(-xe^{\frac{1}{x}}) &= -e^{\frac{1}{x}} - xe^{-\frac{1}{x}}\left(-\frac{1}{x^2}\right) \end{aligned} $$ Now, plug the derivatives into the limit: $$ \begin{aligned} \lim_{x\to 0} g'(x) &= \frac{3-0+4}{2-1} = \frac{7}{1} = 7 \end{aligned} $$ Since the limit exists, we can proceed to analyze the continuity of the function.

Step 2: Determine continuity at x=0

As we found in Step 1, the limit of \(f(x)\) as x approaches 0 is 7. Since the given function's value at x=0 is 0, which is not equal to the calculated limit, the function is not continuous at x=0.

Step 3: Determine differentiability at x=0

Since the function is not continuous at x=0, it is automatically not differentiable at x=0 as well. In conclusion, the function has a limit at x=0, but it is not continuous or differentiable at x=0.

Key Concepts

L'Hôpital's Rule

Understanding L'Hôpital's Rule is instrumental when tackling limits involving indeterminate forms like \( 0/0 \) and \( \infty/\infty \). This rule tells us that, under certain conditions, the limit of a quotient of two functions as they approach a point can be found by taking the limit of the quotient of their derivatives.

In the case of our function, we find that directly substituting \( x = 0 \) into \( g(x) \) gives an indeterminate form \( 0/0 \). Therefore, to solve this, we take the derivative of the numerator and the denominator. The meticulous application of L'Hôpital's Rule requires us to ensure the derivatives exist around the point of interest and that the limit of the derivative quotient exists as well.

Continuing with the given function, when we apply L'Hôpital's Rule and evaluate the limit, the outcome converges to 7, indicating that the rule provides a concrete answer in this scenario. However, this is only part of the full picture when considering the behavior of the function with regards to continuity and differentiability at \( x = 0 \).

Differentiability

Differentiability of a function at a point, simply put, means that the function has a definite and unique tangent at that point, which also translates to having a unique derivative. A deeper dive into the concept reveals that for a function to be differentiable at some point \( x = c \), the function must not only be continuous at that point but also smooth; meaning it doesn't have any sharp corners or cusps.

Taking our given function \( f(x) \), our assessment shows it is not continuous at \( x = 0 \) because the limit of \( f(x) \) as \( x \) approaches 0 is not equal to \( f(0) \). Because continuity is a prerequisite for differentiability, \( f(x) \) fails to be differentiable at \( x = 0 \) as well. It's like trying to draw a straight line on a graph without lifting your pencil, but there's an abrupt stop: you can’t continue smoothly, so there's no derivative at that point.

Indeterminate Forms

Indeterminate forms are expressions that do not have an immediately obvious limit. The most common examples include \( 0/0 \), \( \infty/\infty \), \( 0\times\infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \). These forms are quite troublesome as they can potentially represent a range of different values, or in some cases, no value at all!

To handle these, we often need additional tools - like L'Hôpital's Rule or algebraic manipulation - to discover the true behavior as we approach the limit. In the exercise provided, we encountered the \( 0/0 \) form, and efficiently used L'Hôpital's Rule to resolve it. This illustrates the rule's power to unlock the mysteries behind indeterminate forms, allowing us to understand and describe the behavior of functions at critical points.