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lim_(x rarr0)f(x)" for "f(x)=(1-e^(x))/(ln(2-e^(x))) Calculating lim_(x rarr a)f(x)

lim⁡x→0f(x) for f(x)=1−exln⁡(2−ex) \lim _{x \rightarrow 0} f(x) \text { for } f(x)=\frac{1-e^{x}}{\ln \left(2-e^{x}\right)} \newlineCalculating lim⁡x→af(x) \lim _{x \rightarrow a} f(x)

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Q. lim⁡x→0f(x) for f(x)=1−exln⁡(2−ex) \lim _{x \rightarrow 0} f(x) \text { for } f(x)=\frac{1-e^{x}}{\ln \left(2-e^{x}\right)} \newlineCalculating lim⁡x→af(x) \lim _{x \rightarrow a} f(x)
  1. Identify Limit Expression: Identify the limit expression. lim⁡x→01−exln⁡(2−ex) \lim_{{x \to 0}} \frac{{1 - e^x}}{{\ln(2 - e^x)}}
  2. Substitute x=0x = 0: Substitute x=0x = 0 into the expression. (1−e0)/extln(2−e0)=(1−1)/extln(2−1)=0/extln(1)(1 - e^0) / ext{ln}(2 - e^0) = (1 - 1) / ext{ln}(2 - 1) = 0 / ext{ln}(1)
  3. Simplify Expression: Simplify the expression. 0/00 / 0 is an indeterminate form, so we need to use L'Hôpital's Rule.
  4. Apply L'Hôpital's Rule: Apply L'Hôpital's Rule. Take the derivative of the numerator and the denominator. Numerator: ddx(1−ex)=−ex\frac{d}{dx} (1 - e^x) = -e^x Denominator: ddxln⁡(2−ex)=−ex2−ex\frac{d}{dx} \ln(2 - e^x) = \frac{-e^x}{2 - e^x}
  5. Rewrite Using Derivatives: Rewrite the limit using the derivatives. lim⁡x→0−ex−ex2−ex=lim⁡x→0(2−ex) \lim_{x \to 0} \frac{-e^x}{\frac{-e^x}{2 - e^x}} = \lim_{x \to 0} (2 - e^x)
  6. Substitute x=0x = 0: Substitute x=0x = 0 into the simplified expression. 2−e0=2−1=12 - e^0 = 2 - 1 = 1

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