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f(x)={[x^(2)," for "x <= 0],[ln(x)," for "x > 0]:} Find lim_(x rarr1)f(x). Choose 1 answer: (A) 0 (B) 1 (c) e (D) The limit doesn't exist.

f(x)={x2 for x≤0ln⁡(x) for x>0 f(x)=\left\{\begin{array}{ll} x^{2} & \text { for } x \leq 0 \\ \ln (x) & \text { for } x>0 \end{array}\right. \newlineFind lim⁡x→1f(x) \lim _{x \rightarrow 1} f(x) .\newlineChoose 11 answer:\newline(A) 00\newline(B) 11\newline(C) e e \newline(D) The limit doesn't exist.

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Q. f(x)={x2 for x≤0ln⁡(x) for x>0 f(x)=\left\{\begin{array}{ll} x^{2} & \text { for } x \leq 0 \\ \ln (x) & \text { for } x>0 \end{array}\right. \newlineFind lim⁡x→1f(x) \lim _{x \rightarrow 1} f(x) .\newlineChoose 11 answer:\newline(A) 00\newline(B) 11\newline(C) e e \newline(D) The limit doesn't exist.
  1. Given piecewise function: We are given a piecewise function f(x)f(x) defined as:\newlinef(x)={x2for x≤0, ln⁡(x)for x>0f(x) = \begin{cases} x^2 & \text{for } x \leq 0,\ \ln(x) & \text{for } x > 0 \end{cases}\newlineWe need to find the limit of f(x)f(x) as xx approaches 11.\newlineSince 11 is greater than 00, we will use the definition of f(x)f(x) for x>0x > 0, which is f(x)=ln⁡(x)f(x) = \ln(x).
  2. Limit definition: To find the limit as xx approaches 11 for the natural logarithm function, we substitute xx with 11 in the function ln⁡(x)\ln(x):lim⁡x→1ln⁡(x)=ln⁡(1)\lim_{x \to 1} \ln(x) = \ln(1)
  3. Substitute xx with 11: We know that the natural logarithm of 11 is 00:ln⁡(1)=0\ln(1) = 0Therefore, the limit of f(x)f(x) as xx approaches 11 is 00.

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