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Find the limit as x approaches 0 of (e^x - 1 - x)/x^2

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Answer
12\boxed{\frac{1}{2}}

Checking the result using the Taylor series expansion for exe^x, (1+x+x22+… )−1−xx2=x22+…x2\frac{(1 + x + \frac{x^2}{2} + \dots) - 1 - x}{x^2} = \frac{\frac{x^2}{2} + \dots}{x^2}, which approaches 12\frac{1}{2} as x→0x \to 0.

Problem: Find the limit lim⁡x→0ex−1−xx2\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}.

Steps

  1. Identify indeterminate form
    e0−1−002=00\frac{e^0 - 1 - 0}{0^2} = \frac{0}{0}
  2. Apply L'Hopital's Rule
    lim⁡x→0ddx(ex−1−x)ddx(x2)=lim⁡x→0ex−12x\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1 - x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{e^x - 1}{2x}
  3. Identify indeterminate form again
    e0−12(0)=00\frac{e^0 - 1}{2(0)} = \frac{0}{0}
  4. Apply L'Hopital's Rule a second time
    lim⁡x→0ddx(ex−1)ddx(2x)=lim⁡x→0ex2\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{e^x}{2}
  5. Evaluate the limit by direct substitution
    e02=12\frac{e^0}{2} = \frac{1}{2}

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