Why does e^(-1/x) go to 0 as x → 0^(+)?

Because as x → 0 through positive values, 1/x → +∞, so -1/x → -∞. e^(-∞) = lim as u → ∞ of 1/(e^u) = 0.

JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Limit of the Solution to a Separable First-Order Differential Equation

Let y = f(x) be the solution of the differential equation y(x + 1)dx - x^2dy = 0, y(1) = e. Then lim as x → 0^(+) of f(x) is equal to

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A1/(e^2)Be^2C0D1/e

Step-by-step solutionView

Correct answer

Separating variables yields $y = x e^{2 - 1 / x}$ , which approaches $0$ as $x o 0^{+}$ .

Option analysis

Why each option works or fails

A · 1/(e^2)

Believing that the constant of integration cancels the negative exponent entirely or dropping the factor of $x$ and taking $e^{- 2}$ . Retain the factor of $x$ from the integration $\int \frac{1}{x} d x = \ln ∣ x ∣$ and correctly evaluate the indeterminate form $x e^{- 1 / x}$ as $x \to 0^{+}$ .

B · e^2

Focusing solely on the integration constant $C = e^{2}$ and assuming the limit is determined only by this multiplier. Evaluate the limit of the entire expression $f (x) = x e^{2 - 1 / x}$ as $x \to 0^{+}$ , not just the constant factor $e^{C}$ .

C · 0

The differential equation is solved to find $f (x) = x e^{2} e^{- 1 / x}$ , where both $x \to 0$ and $e^{- 1 / x} \to 0$ as $x \to 0^{+}$ , resulting in $0$ . This option is correct.

D · 1/e

Making a sign error during the integration of $x^{- 2}$ or solving for $C$ , such as writing $\int x^{- 2} d x = + 1 / x$ , leading to an exponent that does not vanish. Recall that $\int x^{- 2} d x = - \frac{1}{x}$ , so the exponent is $- 1 / x$ , which tends to $- \infty$ as $x \to 0^{+}$ .

Reviewed route

Solution

StepWorking

01Given

The differential equation is $y (x + 1) d x - x^{2} d y = 0$ with initial condition $y (1) = e$ , for $x > 0$ .

02Goal

Find the explicit function $y = f (x)$ and evaluate $\lim_{x \to 0^{+}} f (x)$ .

03Approach

Separate variables to write $\frac{d y}{y} = \frac{x + 1}{x^{2}} d x$ . Integrate both sides, find the integration constant using $y (1) = e$ , express $y$ as $f (x)$ , then evaluate the limit as $x \to 0^{+}$ .

04Execute

Separate variables and integrate: $\int \frac{d y}{y} = \int (\frac{1}{x} + \frac{1}{x^{2}}) d x$ $\ln y = \ln x - \frac{1}{x} + C$ Using $y (1) = e$ : $\ln e = \ln 1 - 1 + C ⟹ 1 = - 1 + C ⟹ C = 2$

05Execute

Rewrite the equation explicitly for $y$ : $\ln y - \ln x = 2 - \frac{1}{x} ⟹ \ln (\frac{y}{x}) = 2 - \frac{1}{x}$ $y = x e^{2 - 1 / x}$

06Execute

Evaluate the limit as $x \to 0^{+}$ : $\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} x e^{2} e^{- 1 / x}$ As $x \to 0^{+}$ , $\frac{1}{x} \to + \infty$ , so $e^{- 1 / x} \to 0$ . Therefore, the limit is $0 \cdot e^{2} \cdot 0 = 0$ .

✓Verify

Let $t = 1 / x \to \infty$ . Then $\lim_{t \to \infty} \frac{e^{2}}{t e^{t}} = \frac{e^{2}}{\infty} = 0$ . Both terms ( $x \to 0$ and $e^{- 1 / x} \to 0$ ) push the product decisively to $0$ , confirming consistency.

Hints that build this answer step by step

How does the differential equation $y (x + 1) d x - x^{2} d y = 0$ separate into variables?
$\frac{1}{y} d y = \frac{x + 1}{x^{2}} d x = (\frac{1}{x} + \frac{1}{x^{2}}) d x$
Integrating both sides and applying $y (1) = e$ , what is the explicit formula for $y = f (x)$ ?
$f (x) = x e^{2 - 1 / x}$
What is the value of $\lim_{x \to 0^{+}} x e^{2 - 1 / x}$ ?
$0$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why does

e^{- 1 / x}

go to 0 as

x \to 0^{+}

Because as $x \to 0$ through positive values, $1 / x \to + \infty$ , so $- 1 / x \to - \infty$ . $e^{- \infty} = \lim_{u \to \infty} \frac{1}{e^{u}} = 0$ .

Answer

Separating variables yields $y = x e^{2 - 1 / x}$ , which approaches $0$ as $x o 0^{+}$ .

Why each option works or fails

A: 1/(e^2) - Believing that the constant of integration cancels the negative exponent entirely or dropping the factor of $x$ and taking $e^{- 2}$ . Retain the factor of $x$ from the integration $y (x + 1) d x - x^{2} d y = 0$ and correctly evaluate the indeterminate form $x e^{- 1 / x}$ as $x \to 0^{+}$ .
B: e^2 - Focusing solely on the integration constant $C = e^{2}$ and assuming the limit is determined only by this multiplier. Evaluate the limit of the entire expression $f (x) = x e^{2 - 1 / x}$ as $x \to 0^{+}$ , not just the constant factor $e^{C}$ .
C · correct: 0 - The differential equation is solved to find $f (x) = x e^{2} e^{- 1 / x}$ , where both $x \to 0$ and $e^{- 1 / x} \to 0$ as $x \to 0^{+}$ , resulting in $y = x e^{2 - 1 / x}$ . This option is correct.
D: 1/e - Making a sign error during the integration of $x^{- 2}$ or solving for $C$ , such as writing $\int x^{- 2} d x = + 1 / x$ , leading to an exponent that does not vanish. Recall that $\int x^{- 2} d x = - \frac{1}{x}$ , so the exponent is $- 1 / x$ , which tends to $- \infty$ as $x \to 0^{+}$ .

Step-by-step solution

Given: The differential equation is $y (x + 1) d x - x^{2} d y = 0$ with initial condition $y (1) = e$ , for $x > 0$ .
Goal: Find the explicit function $y = f (x)$ and evaluate $\lim_{x \to 0^{+}} f (x)$ .
Approach: Separate variables to write $\frac{d y}{y} = \frac{x + 1}{x^{2}} d x$ . Integrate both sides, find the integration constant using $y (1) = e$ , express $y$ as $f (x)$ , then evaluate the limit as $x \to 0^{+}$ .
Execute: Separate variables and integrate: $\int \frac{d y}{y} = \int (\frac{1}{x} + \frac{1}{x^{2}}) d x$ $\ln y = \ln x - \frac{1}{x} + C$ Using $y (1) = e$ : $\ln e = \ln 1 - 1 + C ⟹ 1 = - 1 + C ⟹ C = 2$
Execute: Rewrite the equation explicitly for $y$ : $\ln y - \ln x = 2 - \frac{1}{x} ⟹ \ln (\frac{y}{x}) = 2 - \frac{1}{x}$ $y = x e^{2 - 1 / x}$
Execute: Evaluate the limit as $x \to 0^{+}$ : $\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} x e^{2} e^{- 1 / x}$ As $x \to 0^{+}$ , $\frac{1}{x} \to + \infty$ , so $e^{- 1 / x} \to 0$ . Therefore, the limit is $0 \cdot e^{2} \cdot 0 = 0$ .
Verify: Let $t = 1 / x \to \infty$ . Then $\lim_{t \to \infty} \frac{e^{2}}{t e^{t}} = \frac{e^{2}}{\infty} = 0$ . Both terms ( $x \to 0$ and $e^{- 1 / x} \to 0$ ) push the product decisively to $0$ , confirming consistency.

Limit of the Solution to a Separable First-Order Differential Equation

What feels right?

Separating variables yields $y = x e^{2 - 1 / x}$ , which approaches $0$ as $x o 0^{+}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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