Problem 172 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt[4]{1+2 x}}{x} \text { \\{Ans. }-\frac{1}{6}\right\\} $$

Step-by-Step Solution

Answer

The limit of the given expression as x approaches 0 is \(-\frac{1}{6}\).

Step 1: Identify indeterminate form

First, let's try plugging in x=0 into the given expression: \[ \frac{\sqrt[3]{1+0}-\sqrt[4]{1+2(0)}}{0} = \frac{1-1}{0}, \] which is an indeterminate form (0/0). So we need to further simplify the expression to find the limit.

Step 2: Apply the L'Hospital's rule

L'Hospital's Rule states that if the limit of a function has the form 0/0 or ∞/∞, then: \[ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}, \] provided that the limit on the right-hand side exists. In this case, we need to find the derivatives of the functions in the numerator and denominator.

Step 3: Find the derivatives of the numerator And denominator

We will find the derivative of the numerator with respect to x: \[ \frac{d}{dx} (\sqrt[3]{1+x}-\sqrt[4]{1+2 x}) = \frac{d}{dx} (\sqrt[3]{1+x}) - \frac{d}{dx} (\sqrt[4]{1+2 x}). \] Now, finding the individual derivatives: \[ \frac{d}{dx} (\sqrt[3]{1+x}) = \frac{1}{3}(1+x)^{-\frac{2}{3}}, \] and \[ \frac{d}{dx} (\sqrt[4]{1+2 x}) = \frac{1}{2}(1+2 x)^{-\frac{3}{4}}. \] The derivative of the denominator is simply 1.

Step 4: Apply L'Hospital's Rule

Now, we apply L'Hospital's Rule: \[ \lim_{x \rightarrow 0} \frac{\sqrt[3]{1+x}-\sqrt[4]{1+2 x}}{x} = \lim_{x \rightarrow 0} \frac{\frac{1}{3}(1+x)^{-\frac{2}{3}} - \frac{1}{2}(1+2 x)^{-\frac{3}{4}}}{1}. \]

Step 5: Evaluate the limit

Substitute x=0 into the simplified expression: \[ \lim_{x \rightarrow 0} \frac{\frac{1}{3}(1+x)^{-\frac{2}{3}} - \frac{1}{2}(1+2 x)^{-\frac{3}{4}}}{1} = \frac{1}{3}(1)^{-\frac{2}{3}} - \frac{1}{2}(1)^{-\frac{3}{4}} = \frac{1}{3}-\frac{1}{2}=-\frac{1}{6}. \] Therefore, the limit of the given expression as x approaches 0 is -1/6.

Key Concepts

L'Hospital's Rule

L'Hospital's Rule is a powerful technique in calculus for finding limits.
It is particularly useful when you encounter indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
Here's how it works:

• When a limit problem yields an indeterminate form, apply L'Hospital's Rule by differentiating the numerator and denominator separately.
• Then, take the limit again of the new fraction.
• If the result is still indeterminate, you might need to apply the rule more than once.

For the original exercise, the initial substitution of \( x = 0 \) resulted in \( \frac{0}{0} \), so L'Hospital's Rule was applicable. By taking the derivatives of both the numerator and denominator, and evaluating again, the limit was successfully found.

Indeterminate Form

In calculus, an indeterminate form is an expression that doesn't immediately provide a clear limit.
Common examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \(0 \cdot \infty\), and others.
These forms indicate ambiguity, as they don't conform to normal arithmetic rules.

• When encountering an indeterminate form, it's a signal that the limit requires further work.
• Techniques like L'Hospital's Rule, algebraic manipulation, or series expansion can resolve these forms.

In the original solution, substituting \( x = 0 \) gave us \( \frac{0}{0} \), a classic indeterminate form. This flagged the need for more sophisticated methods to find the actual limit.

Limit of a Function

The limit of a function describes the behavior of that function as it approaches a particular point.
Limits can tell you how a function behaves near a point without having to actually evaluate the function at that point.
They are fundamental in defining derivatives and integrals.

• A limit is expressed as \( \lim_{x \to a} f(x) \), representing the value that \( f(x) \) approaches as \( x \) approaches \( a \).
• Limits can approach finite numbers or infinity, but sometimes they might not exist at all.

In our exercise, we sought the limit as \( x \to 0 \). By applying L'Hospital's Rule to address the indeterminate form, we accurately evaluated the function's behavior near zero, finding the precise value of \(-\frac{1}{6}\).