Problem 412 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

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

Step-by-Step Solution

Answer

The short answer to the limit question is: \[ \lim _{x \rightarrow 0} \frac{e^{-\frac{1}{x}}}{x^{a}} = 0 \]

Step 1: Check the indeterminate form

Let's first see if the expression is in an indeterminate form when x approaches 0: \[ \lim _{x \rightarrow 0} \frac{e^{-\frac{1}{x}}}{x^{a}} \] As x approaches 0, the exponent of e approaches -∞ because -1/x approaches -∞. So, e raised to this very large negative number will approach 0. The denominator, x^a, also approaches 0 as x approaches 0. Our function is, therefore, of the indeterminate form 0/0, and we will apply L'Hopital's rule to solve this limit.

Step 2: Differentiate the numerator and denominator

Applying L'Hopital's Rule, we differentiate the numerator and denominator of the expression with respect to x: \[ \frac{d}{dx} \left( e^{-\frac{1}{x}}\right) = \frac{1}{x^2} e^{-\frac{1}{x}} \] \[ \frac{d}{dx} \left( x^a \right) = a x^{a-1} \]

Step 3: Find the limit of the derivative

Now substitute these derivatives back into the expression and find the limit as x approaches 0: \[ \lim _{x \rightarrow 0} \frac{\frac{1}{x^2} e^{-\frac{1}{x}}}{a x^{a-1}} \] This limit is now of the form 0/0, an indeterminate form again. We will apply L'Hopital's rule one more time to solve the limit.

Step 4: Differentiate once more

Differentiate the numerator and denominator of the new limit with respect to x: \[ \frac{d}{dx} \left( \frac{1}{x^2} e^{-\frac{1}{x}} \right) = -\frac{2}{x^3} e^{-\frac{1}{x}} + \frac{1}{x^4} e^{-\frac{1}{x}} \] \[ \frac{d}{dx} \left( a x^{a-1} \right) = a(a-1)x^{a-2} \]

Step 5: Find the final limit

Now substitute these derivatives back into the expression and find the limit as x approaches 0: \[ \lim _{x \rightarrow 0} \frac{-\frac{2}{x^3} e^{-\frac{1}{x}} + \frac{1}{x^4} e^{-\frac{1}{x}}}{a(a-1)x^{a-2}} \] When x approaches 0, the terms with x in the denominator will approach ∞. However, the exponential term in the numerator approaches 0 faster than the power function. This implies that the limit of the whole expression is 0. The limit is: \[ \lim _{x \rightarrow 0} \frac{e^{-\frac{1}{x}}}{x^{a}} = 0 \]

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus for evaluating the limits of indeterminate forms. An indeterminate form occurs when substituting the limit point into a function results in an expression like 0/0 or ∞/∞. When you're confronted with such a limit, L'Hopital's Rule permits taking the derivative of the numerator and the derivative of the denominator and then re-evaluating the limit of that new ratio.

The rule is not a catch-all solution and can only be used under certain conditions. First, the limit must be indeterminate. Second, the derivatives of the numerator and denominator must exist, and finally, the limit of the derivatives must be determinable. You may sometimes need to apply L'Hopital's Rule multiple times before reaching a solvable limit, as seen in the textbook example.

Indeterminate Forms

Indeterminate forms occur when limits yield an expression that is not immediately clear in determining the limit's value. The most common indeterminate forms are 0/0, ∞/∞, 0*∞, ∞ - ∞, 1^∞, 0^0, and ∞^0. In our exercise, we encountered the 0/0 form, where both the numerator and the denominator approach zero as x approaches zero. To solve these, we often have to manipulate the expressions (for example, through factoring, conjugation, or using L'Hopital's Rule) until the limit can be directly calculated or further analyzed.

Exponential Functions Differentiation

Exponential functions are characterized by the variable appearing in the exponent. Differentiating exponential functions often involves the chain rule, especially when the exponent is more complex than a simple x. The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

In the given exercise, the function in the numerator is of the form \( e^{-\frac{1}{x}} \), which is an exponential function with a chain rule situation because of the \( -\frac{1}{x} \) in the exponent. The differentiation results in a factor that originates from the exponent's derivative, which is crucial to solving the limit.

Limits Approaching Zero

In calculus, evaluating limits as x approaches zero can be especially tricky if the function involves terms that become infinite or indeterminate at zero. As seen in the exercise, the function \( \frac{e^{-\frac{1}{x}}}{x^{a}} \) is not straightforward when x is nearing zero. Understanding the behavior of functions near zero is critical to mastering limits.

While some functions will clearly approach a specific value, others may oscillate or grow without bound, and still, others will take on an indeterminate form that requires further analysis like L'Hopital's Rule or algebraic manipulation to resolve. The key lies in recognizing patterns and understanding the properties of functions, such as exponential decay or growth and the behavior of polynomial functions near zero.