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Practice Problems for 2/6

  1. Evaluate the limit and justify each step by citing the appropriate Limit Laws. \[ \lim_{t\rightarrow 2}(t+1)^9(\sqrt{t^2-1}). \]

    2. -Product rule with \(f(t)= (t+1)^9\) and \(g(t)=\sqrt{t^2-1}\).
    -For \(\lim_{t\rightarrow 2} f(t)\) we'll technically first apply the addition rule, then apply the power rule
    -For \(\lim_{t\rightarrow 2} g(t)\) we'll apply the power rule (for \(t^2\)), followed by the difference rule, then of course the root rule.
    This gives \(\lim_{t\rightarrow 2}(t+1)^9(\sqrt{t^2-1})=3^9\sqrt{3}=3^{19/2}\).
  2. If all you know is that \(3x \leq f(x) \leq x^3+x^2+1\) for all \(x\), what can you say about \(\lim_{x\rightarrow 1} f(x)\)?

    3. Since \(\lim_{x \rightarrow 1} 3x=3\) and \( \lim_{x \rightarrow 1} x^3+x^2+1=3\), it follows that \[ 3 \leq \lim_{x \rightarrow 1} f(x) \leq 3, \] so \(\lim_{x \rightarrow 1} f(x)=3\). (This is the Sandwich Theorem.)
  3. Since the absolute value is involved, we should use one-sided limits. \[ \lim_{x\rightarrow 0^+}\frac{|x|}{x}=\lim_{x\rightarrow 0^+}\frac{x}{x}=\lim_{x\rightarrow 0^+}1=1. \] Also, \[ \lim_{x\rightarrow 0^-}\frac{|x|}{x}=\lim_{x\rightarrow 0^-}\frac{-x}{x}=\lim_{x\rightarrow 0^-}-1=-1. \] Since the right and left limits do not agree, we can conclude that the limit does not exist.