Limits: When Does the Substitution Rule Work?
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Problems 5
- Evaluate limx→−2−{−3xx≤−21−x3−2<x<2(x−3)2−8x≥2\lim_{x\to-2^-}{\begin{cases}-3x&x\leq-2\\1-x^3&-2<x<2\\(x-3)^2-8&x\geq2\end{cases}}x→−2−lim⎩⎨⎧−3x1−x3(x−3)2−8x≤−2−2<x<2x≥2
- Evaluate limx→−2+{−3xx≤−21−x3−2<x<2(x−3)2−8x≥2\lim_{x\to-2^+}{\begin{cases}-3x&x\leq-2\\1-x^3&-2<x<2\\(x-3)^2-8&x\geq2\end{cases}}x→−2+lim⎩⎨⎧−3x1−x3(x−3)2−8x≤−2−2<x<2x≥2
- Evaluate limx→2−{−3xx≤−21−x3−2<x<2(x−3)2−8x≥2\lim_{x\to2^-}{\begin{cases}-3x&x\leq-2\\1-x^3&-2<x<2\\(x-3)^2-8&x\geq2\end{cases}}x→2−lim⎩⎨⎧−3x1−x3(x−3)2−8x≤−2−2<x<2x≥2
- Evaluate limx→2+{−3xx≤−21−x3−2<x<2(x−3)2−8x≥2\lim_{x\to2^+}{\begin{cases}-3x&x\leq-2\\1-x^3&-2<x<2\\(x-3)^2-8&x\geq2\end{cases}}x→2+lim⎩⎨⎧−3x1−x3(x−3)2−8x≤−2−2<x<2x≥2
- Find the limit of the following function using substitution: \\ limx→−4x2+x−12x+4=\lim\limits_{x \to -4} \frac{x^2+x-12}{x+4}=x→−4limx+4x2+x−12=