LimitseasyFree

Limits — JEE Maths practice question

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

Question

\displaystyle\lim_{x\to 0}\dfrac{ae^{x}+b\cos x+ce^{-x}}{e^{2x}-2e^{x}+1}=4

. Then

a=2

correct

b=-4

correct

c=2

correct

a+b+c=-8

Solution

Step 1: Examine the denominator first:

e^{2x}-2e^{x}+1=(e^{x}-1)^{2}\sim x^{2}\text{ as }x\to 0

So the denominator behaves like

x^{2}

near

0

. Step 2: For the limit to be finite, the numerator must also go to

0

. At

x=0

a+b+c=0\quad\ldots(i)

Step 3: Apply L'Hôpital's rule once. Differentiate numerator and denominator:

\lim_{x\to 0}\dfrac{ae^{x}-b\sin x-ce^{-x}}{2e^{2x}-2e^{x}}

The denominator at

x=0

2-2=0

, so for the limit to exist, the numerator must also be

0

x=0

a-c=0\;\Rightarrow\;a=c\quad\ldots(ii)

Step 4: Apply L'Hôpital's rule again:

\lim_{x\to 0}\dfrac{ae^{x}-b\cos x+ce^{-x}}{4e^{2x}-2e^{x}}

x=0

: numerator

=a-b+c

, denominator

=4-2=2

. Step 5: Set equal to

4

\dfrac{a-b+c}{2}=4\;\Rightarrow\;a-b+c=8\quad\ldots(iii)

Step 6: Solve the three equations

(i),(ii),(iii)

: From

(i)

and

(iii)

: adding,

2(a+c)=8\;\Rightarrow\;a+c=4

, and from

(i)

b=-(a+c)=-4

. From

(ii)

and

a+c=4

2a=4\;\Rightarrow\;a=2

, and

c=2

. Step 7: Check sum:

a+b+c=2+(-4)+2=0

, not

-8

. So (D) is wrong. Correct answers: (1), (2) 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.