Why does go to 0 as ?

Because as through positive values, , so . .

JEE MainMathematics

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Differential Equations: Mathematics | JEE Main

Let

y = f(x)

be the solution of the differential equation

y(x + 1)dx - x^2dy = 0, y(1) = e

. Then

\lim_{x \to 0^+} f(x)

is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\frac{1}{e^2}

e^2

0

\frac{1}{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 ·

\frac{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 $\int \frac{1}{x} dx = \ln|x|$ 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

None. 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 ·

\frac{1}{e}

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

Reviewed route

Solution

StepWorking

01given

The differential equation is $y(x + 1)dx - x^2dy = 0$ 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 $\frac{dy}{y} = \frac{x+1}{x^2}dx$ . 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: $\int \frac{dy}{y} = \int \left(\frac{1}{x} + \frac{1}{x^2}\right) dx$ $\ln y = \ln x - \frac{1}{x} + C$ Using $y(1) = e$ : $\ln e = \ln 1 - 1 + C \implies 1 = -1 + C \implies C = 2$

05execute

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

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 $y(x + 1)dx - x^2dy = 0$ separate into variables?
$\frac{1}{y} dy = \frac{x + 1}{x^2} dx = \left(\frac{1}{x} + \frac{1}{x^2}\right) dx$
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
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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: $\frac{1}{e^2}$ - 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 $\int \frac{1}{x} dx = \ln|x|$ 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 · correct: $0$ - None. 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: $\frac{1}{e}$ - Making a sign error during the integration of $x^{-2}$ or solving for $C$ , such as writing $\int x^{-2}dx = +1/x$ , leading to an exponent that does not vanish. Recall that $\int x^{-2} dx = -\frac{1}{x}$ , so the exponent is $-1/x$ , which tends to $-\infty$ as $x \to 0^+$ .

Step-by-step solution

given: The differential equation is $y(x + 1)dx - x^2dy = 0$ 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 $\frac{dy}{y} = \frac{x+1}{x^2}dx$ . 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: $\int \frac{dy}{y} = \int \left(\frac{1}{x} + \frac{1}{x^2}\right) dx$ $\ln y = \ln x - \frac{1}{x} + C$ Using $y(1) = e$ : $\ln e = \ln 1 - 1 + C \implies 1 = -1 + C \implies C = 2$
execute: Rewrite the equation explicitly for $y$ : $\ln y - \ln x = 2 - \frac{1}{x} \implies \ln\left(\frac{y}{x}\right) = 2 - \frac{1}{x}$ $y = x e^{2 - 1/x}$
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.

Differential Equations: Mathematics | JEE Main

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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