LimitseasyFree

Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{x\to 0}g(x)

exists and

\displaystyle\lim_{x\to 0}g(x)\cdot\left(\dfrac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}\right)

also exists, then

g(x)

is(are) equal to

x

correct

x+4

x^{2}

correct

4x^{2}+3x+2

Solution

Step 1: Let

f(x)=\dfrac{e^{1/x}-e^{-1/x}}{e^{1/x}+e^{-1/x}}

. Step 2: Compute the LHL. As

x\to 0^{-}

\dfrac{1}{x}\to -\infty

, so

e^{1/x}\to 0

while

e^{-1/x}\to\infty

. Multiply numerator and denominator by

e^{1/x}

f(x)=\dfrac{e^{2/x}-1}{e^{2/x}+1}\;\to\;\dfrac{0-1}{0+1}=-1

Step 3: Compute the RHL. As

x\to 0^{+}

\dfrac{1}{x}\to +\infty

, so

e^{-1/x}\to 0

while

e^{1/x}\to\infty

. Multiply numerator and denominator by

e^{-1/x}

f(x)=\dfrac{1-e^{-2/x}}{1+e^{-2/x}}\;\to\;\dfrac{1-0}{1+0}=1

Step 4: So LHL

=-1

and RHL

=1

, and

f(x)

has a jump discontinuity at

x=0

. Step 5: For

\displaystyle\lim_{x\to 0}g(x)\cdot f(x)

to exist, we need

\lim_{x\to 0^{-}}g(x)\cdot(-1)=\lim_{x\to 0^{+}}g(x)\cdot 1

Since

\displaystyle\lim_{x\to 0}g(x)

exists (call it

L

), this becomes

-L=L\;\Rightarrow\;L=0

. Step 6: So we need

g(0)=0

(i.e.,

g

vanishes at

0

). Check options: (A)

g(x)=x

g(0)=0

. Valid. (B)

g(x)=x+4

g(0)=4\neq 0

. Invalid. (C)

g(x)=x^{2}

g(0)=0

. Valid. (D)

g(x)=4x^{2}+3x+2

g(0)=2\neq 0

. Invalid. Correct answers: (1) and (3)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.