If a function is defined by f(x) = begincases 1 + x, & x leq -1 x2, & | Mathematics I 2079

Mathematics I207910 marksNumerical: piecewise function limitsPrecise definition of Limit

If a function is defined by $f(x) = \begin{cases} 1 + x, & x \leq -1 \\ x^2, & x > -1 \end{cases}$ , evaluate $f(-3)$ , $f(-1)$ and $f(0)$ and sketch the graph. Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. [10+0]

If a function is defined by $f(x) = \begin{cases} 1 + x, & x \leq -1 \ x^2, & x > -1 \end{cases}$, evaluate $f(-3)$, $f(-1)$ and $f(0)$ and sketch the graph. Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. [10+0]

$f(x) = \begin{cases} 1 + x, & x \leq -1 \\ x^2, & x -1 \end{cases}$ Evaluate $f(-3)$ , $f(-1)$ , $f(0)$ ; sketch graph; prove $\lim{x\to 0}\frac{x}{x}$ does not exist. --- $f(-3)$ : Since $-3 \leq -1$ , use $1+x$ : $f(-3) = 1 + (-3) = -2$ $f(-1)$ : Since $-1 \leq -1$ , use $1+x$ : $$f(-1) = 1 + (-1) ...

$$f(x) = \begin{cases} 1 + x, & x \leq -1 \ x^2, & x -1 \end{cases}$$ Evaluate $f(-3)$, $f(-1)$, $f(0)$; sketch graph; prove $\lim{x\to 0}\frac{x}{x}$ does not exist. --- $f(-3)$: Since $-3 \leq -1$, use $1+x$: $$f(-3) = 1 + (-3) = -2$$ $f(-1)$: Since $-1 \leq -1$, use $1+x$: $$f(-1) = 1 + (-1) ...

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If a function is defined by f(x) = begincases 1 + x, & x leq -1 x2, & x > -1 endcases, evaluate f(-3), f(-1) and f(0) and sketch the graph. Prove that limx to 0 frac|x|x does not exist. [10+0]