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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)={x2amp; for x0ln(x)amp; for xgt;0 f(x)=\left\{\begin{array}{ll} x^{2} &amp; \text { for } x \leq 0 \\ \ln (x) &amp; \text { for } x&gt;0 \end{array}\right. \newlineFind limx1f(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.

Full solution

Q. f(x)={x2 for x0ln(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 limx1f(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)={x2amp;for x0, ln(x)amp;for xgt;0f(x) = \begin{cases} x^2 &amp; \text{for } x \leq 0,\ \ln(x) &amp; \text{for } x &gt; 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 > 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):limx1ln(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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