Math  /  Word Problems

QuestionSketch the graph of the piecewise function and find the limits:
1. For f(x)={sinxx<0x20x<2xx2f(x) = \begin{cases} \sin x & x < 0 \\ x^2 & 0 \leq x < 2 \\ x & x \geq 2 \end{cases}, find: i. limx0f(x)\lim_{x \to 0} f(x) ii. limx2f(x)\lim_{x \to 2} f(x)
2. For f(x)={exx0x+10<x<1lnxx1f(x) = \begin{cases} e^x & x \leq 0 \\ |x| + 1 & 0 < x < 1 \\ \ln x & x \geq 1 \end{cases}, find: i. limx0f(x)\lim_{x \to 0} f(x) ii. limx1f(x)\lim_{x \to 1} f(x)

Studdy Solution
Since the function does not approach the same value from both directions, the limit as xx approaches1 does not exist.
In conclusion, for the first piecewise function, limxf(x)=\lim{x \rightarrow} f(x) = and the limit as xx approaches2 does not exist. For the second piecewise function, limxf(x)=1\lim{x \rightarrow} f(x) =1 and the limit as xx approaches1 does not exist.

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