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Limits: Question

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

Question
lim⁡x→0aex+bcos⁡x+ce−xe2x−2ex+1=4\displaystyle\lim_{x\to 0}\dfrac{ae^{x}+b\cos x+ce^{-x}}{e^{2x}-2e^{x}+1}=4. Then
Aa=2a=2correct
Bb=−4b=-4correct
Cc=2c=2correct
Da+b+c=−8a+b+c=-8
Solution
Step 1: Examine the denominator first:
e2x−2ex+1=(ex−1)2∼x2 as x→0e^{2x}-2e^{x}+1=(e^{x}-1)^{2}\sim x^{2}\text{ as }x\to 0
 So the denominator behaves like x2x^{2} near 00. Step 2: For the limit to be finite, the numerator must also go to 00. At x=0x=0:
a+b+c=0…(i)a+b+c=0\quad\ldots(i)
 Step 3: Apply L'Hôpital's rule once. Differentiate numerator and denominator:
lim⁡x→0aex−bsin⁡x−ce−x2e2x−2ex\lim_{x\to 0}\dfrac{ae^{x}-b\sin x-ce^{-x}}{2e^{2x}-2e^{x}}
 The denominator at x=0x=0 is 2−2=02-2=0, so for the limit to exist, the numerator must also be 00 at x=0x=0:
a−c=0  ⇒  a=c…(ii)a-c=0\;\Rightarrow\;a=c\quad\ldots(ii)
 Step 4: Apply L'Hôpital's rule again:
lim⁡x→0aex−bcos⁡x+ce−x4e2x−2ex\lim_{x\to 0}\dfrac{ae^{x}-b\cos x+ce^{-x}}{4e^{2x}-2e^{x}}
 At x=0x=0: numerator =a−b+c=a-b+c, denominator =4−2=2=4-2=2. Step 5: Set equal to 44:
a−b+c2=4  ⇒  a−b+c=8…(iii)\dfrac{a-b+c}{2}=4\;\Rightarrow\;a-b+c=8\quad\ldots(iii)
 Step 6: Solve the three equations (i),(ii),(iii)(i),(ii),(iii): From (i)(i) and (iii)(iii): adding, 2(a+c)=8  ⇒  a+c=42(a+c)=8\;\Rightarrow\;a+c=4, and from (i)(i): b=−(a+c)=−4b=-(a+c)=-4. From (ii)(ii) and a+c=4a+c=4: 2a=4  ⇒  a=22a=4\;\Rightarrow\;a=2, and c=2c=2. Step 7: Check sum: a+b+c=2+(−4)+2=0a+b+c=2+(-4)+2=0, not −8-8. So (D) is wrong. Correct answers: (1), (2) and (3)
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