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Limits: Finite

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

Question
If lim⁡x→0eax−ex−xx2=b\displaystyle\lim_{x\to 0}\dfrac{e^{ax}-e^{x}-x}{x^{2}}=b (finite), then
Aa=2, b=0a=2,\,b=0
Ba=0, b=32a=0,\,b=\tfrac{3}{2}
Ca=2, b=32a=2,\,b=\tfrac{3}{2}correct
Da=0, b=2a=0,\,b=2
Solution
**Step 1: Check the form.** At x=0x=0: numerator =1−1−0=0=1-1-0=0, denominator =0=0. Form is 00\dfrac{0}{0}. **Step 2: Apply L'Hôpital's rule.**
lim⁡x→0aeax−ex−12x\lim_{x\to 0}\dfrac{ae^{ax}-e^{x}-1}{2x}
 **Step 3: Condition for finite limit.** Since denominator →0\to 0, the new numerator must also →0\to 0:
ae0−e0−1=a−1−1=a−2=0  ⇒  a=2ae^{0}-e^{0}-1=a-1-1=a-2=0\;\Rightarrow\; a=2
 **Step 4: Substitute a=2a=2 and apply L'Hôpital again.**
lim⁡x→02e2x−ex−12x  →L’H  lim⁡x→04e2x−ex2\lim_{x\to 0}\dfrac{2e^{2x}-e^{x}-1}{2x}\;\xrightarrow{\text{L'H}}\;\lim_{x\to 0}\dfrac{4e^{2x}-e^{x}}{2}
 **Step 5: Direct substitution.**
=4−12=32=b=\dfrac{4-1}{2}=\dfrac{3}{2}=b
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