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11.21 Limit of function

Tutorial 6.pdf

  1. Use the \(\epsilon\)-\(\delta\) definition of limit to prove the following limits:

    • \(\lim_{x \to a} x^2 = a^2\)

    Let's analyze \(|f(x) - L|\).

    \(|x^2 - a^2| = |(x-a)(x+a)| = |x-a||x+a|\)

    Here is the expression that I need to relate with \(\delta\).

    We need an upper bound for \(|x+a|\). In order to get it, suppose \(\delta < 1\), then

    \(|x-a| < \delta < 1.\)

    This implies: \(|x| - |a| < |x-a| \implies |x| < 1 + |a|.\)

    Thus, \(|x+a| \leq |x| + |a| < 1 + |a| + |a| = 1 + 2|a|.\)

    Hence, \(|x+a| < 1 + 2|a|.\)

    \(|x^2 - a^2| < (1 + 2|a|)|x-a| < \varepsilon\) where \((1 + 2|a|)\) is a constant.

    Therefore \(\delta<\frac{\varepsilon}{1+2|a|}\), then \(\delta = \min\{1, \frac{\varepsilon}{1 + 2|a|}\}\) * \(\lim_{x \to 1} \frac{1}{x} = 1\)

    Analyze: \(\left|\frac{1}{x}-1\right|=\left|\frac{1-x}{x}\right|=\frac{|x-1|}{|x|}\)

    I need an upper bound for \(\frac{1}{|x|}\).

    If \(\delta < 1\), then \(|x-1| < \delta < 1\).

    Let \(|x-1| < 1 \implies |x| - 1 < |x-1| < 1\)

    This implies \(|x| < 2\)

    Then \(-2 < x < 2\) then \(-\frac{1}{2} > \frac{1}{x} > \frac{1}{2}\)

    But \(x\) can be \(0\), problem!

    Let's assume \(\delta < \frac{1}{2}\), \(|x-1| < \delta < \frac{1}{2}\)

    Then \(|x-1| < \frac{1}{2} \implies -\frac{1}{2} < x-1 < \frac{1}{2}\)

    Then \(\frac{1}{|x|} < 2\)

    Back to the problem

    \(\left|\frac{1}{x}-1\right|=\frac{|x-1|}{|x|}<2|x-1|<\varepsilon\)

    Then \(\delta < \frac{\varepsilon}{2}\)

    Finally, \(\delta = \min \left\{ \frac{1}{2}, \frac{\varepsilon}{2} \right\}\)

    Choosing this \(\delta\), the definition holds. * \(\lim_{x \to a} \sqrt{x} = \sqrt{a}, \, a > 0\)

    \(\left| \sqrt{x} - \sqrt{a} \right| = \left| \frac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{\sqrt{x} + \sqrt{a}} \right| = \frac{|x-a|}{\sqrt{x} + \sqrt{a}}\)

    I need an upper bound for \(\frac{1}{\sqrt{x} + \sqrt{a}}\).

    Assume \(\delta < 1\), then \(|x-a| < \delta < 1\).

    Then \(|x|-|a|<|x-a|<1\implies|x|<1+|a|\implies0<|x|<1+|a|.\)

    Then \(0<(\sqrt{x})^2<(\sqrt{1+|a|})^2\implies0<\sqrt{x}<\sqrt{1+|a|}\)

    Then \(\sqrt{a} < \sqrt{x} + \sqrt{a} < \sqrt{1 + |a|} + \sqrt{a}.\)

    Thus \(\frac{1}{\sqrt{a}} > \frac{1}{\sqrt{x} + \sqrt{a}} > \frac{1}{\sqrt{1 + |a|} + \sqrt{a}}\)

    \(\left|\sqrt{x}-\sqrt{a}\right|=\frac{|x-a|}{\sqrt{x}+\sqrt{a}}\leq\frac{|x-a|}{\sqrt{a}}<\varepsilon\) when \(\delta < \sqrt{a}\varepsilon\)

    Choosing \(\delta = \min\{\sqrt{a}\varepsilon, 1\}\)

  2. Compute the following limits using factoring cases:

    • \(\lim_{x \to 2} \frac{x^3 - 8}{x - 2}\)

    \(=\lim_{x\to2}\frac{\left(x-2\right)\left(x^2+2x+4\right)}{x-2}\) since \(x\neq 2\), then \(=\lim_{x\to2}x^2+2x+4=12\)​ * \(\lim_{x \to 1} \frac{2x^2 - 4x + 2}{x - 1}\)

    \(=\lim_{x\to1}\frac{\left(x-1\right)\left(2x-2\right)}{x-1}=2x-2=0\)​ * \(\lim_{x \to 1} \frac{1 - \sqrt{x}}{1 - x}\)

    \(=\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\lim_{x\to1}\frac{1}{1+\sqrt{x}}=\frac12\)​

  3. Prove that \(\lim_{x \to 0} \sin\left(\frac{1}{x}\right)\) does not exist.

    Suppose \(\lim_{x \to 0} \sin\left(\frac{1}{x}\right)=\ell\)

    Then we need to find \(|f(x)-\ell|<\varepsilon\Rightarrow|\sin\left(\frac{1}{x}\right)|<\varepsilon\)​

    Let's \(\delta\), then \(0<|x|<\delta\Rightarrow-\delta<x<\delta\) which means \(x\) can be \(0\)​

    Thus it is a contradiction

  4. Algebra of limits: Let \(f, g\) be real functions and \(x_0\) is a limit point of \(\text{Dom}f\) and \(\text{Dom}g\). Suppose \(\lim_{x\to x_0}f(x)=L_1\) and \(\lim_{x\to x_0}g(x)=L_2\), then:

    • \(\lim_{x \to x_0} cf(x) = c \lim_{x \to x_0} f(x) = cL\)

    Since \(\lim_{x \to x_0} f(x) = L\), then \(\forall\epsilon_1>0,\exists\delta>0\) such that \(0<|x-x_0|<\delta,x\in X\implies|f(x)-\ell|<\epsilon_1\).

    Then we have \(|cf(x)-c\ell|<\epsilon\) if we take \(\varepsilon=\frac{\varepsilon_1}{c}\)​ * \(\lim_{x\to x_0}(f+g)(x)=\lim_{x\to x_0}f(x)+\lim_{x\to x_0}g(x)=L_1+L_2\) * \(\lim_{x\to x_0}(f\cdot g)(x)=\lim_{x\to x_0}f(x)\cdot\lim_{x\to x_0}g(x)=L_1\cdot L_2\)

    \(\lim_{x\to a}f(x)=L_1,\,\forall\frac{\varepsilon}{2\left(1+\left|L_2\right|\right)}>0,\,\exists\delta_1>0\,\text{such that}\,\forall x\in\text{Dom}f\,\text{and}\,0<|x-a|<\delta_1,\,\text{then}\,|f(x)-L_1|<\frac{\varepsilon}{2\left(1+\left|L_2\right|\right)}\)

    \(\lim_{x\to a}g(x)=L_2,\,\frac{\forall\varepsilon}{2\left(1+\left|L_1\right|\right)}>0,\,\exists\delta_2>0\,\text{such that}\,\forall x\in\text{Dom}g\,\text{and}\,0<|x-a|<\delta_2,\,\text{then}\,|g(x)-L_2|<\frac{\varepsilon}{2\left(1+\left|L_1\right|\right)}\)

    We must show that \(\forall \varepsilon > 0, \, \exists \delta > 0 \, \text{such that} \, \forall x \in \text{Dom} f \cap \text{Dom} g \, \text{and} \, 0 < |x-a| < \delta, \, \text{we have} \, |f(x)g(x) - L_1L_2| < \varepsilon.\)

    \(\left| f(x)g(x) - L_1L_2 \right| = \left| f(x)g(x) - L_1g(x) + L_1g(x) - L_1L_2 \right|\)

    \(= \left| g(x)(f(x) - L_1) + L_1(g(x) - L_2) \right|\) \(\leq \left| g(x) \right| \cdot \left| f(x) - L_1 \right| + \left| L_1 \right| \cdot \left| g(x) - L_2 \right|\)

    I need an upper bound for \(g(x)\): \(\left| g(x) \right| = \left| g(x) + L_2 - L_2 \right|\) \(<\left|g(x)-L_2\right|+\left|L_2\right|\) \(<1+\left|L_2\right|\)

    Assume that \(\left| g(x) - L_2 \right| < 1\) and this holds for \(0 < |x-a| < \delta_3\).

    Back to \(\left|f(x)g(x)-L_1L_2\right|\leq\left|g(x)\right|\cdot\left|f(x)-L_1\right|+\left|L_1\right|\cdot\left|g(x)-L_2\right|\)

    \(<(1 + |L_2|)|f(x) - L_1| + |L_1||g(x) - L_2|\)

    \(<(1 + |L_2|)|f(x) - L_1| + (|L_1| + 1)|g(x) - L_2| < \varepsilon\)

    \(\delta = \min\{\delta_1, \delta_2, \delta_3\}\)

    With this \(\delta\), the definition holds. * \(\lim_{x \to x_0} \frac{1}{g(x)} = \frac{1}{M}, \text{ if } M \neq 0\)

  5. Right and Left limits:

    • \(\lim_{x \to a^+} f(x) = L_1 \text{ if for all } \epsilon > 0, \text{ there exists } \delta > 0 \text{ such that for all } x \in \text{Dom}f \cap (a, a + \delta), \text{ we have } |f(x) - L_1| < \epsilon\)
    • \(\lim_{x \to a^-} f(x) = L_2 \text{ if for all } \epsilon > 0, \text{ there exists } \delta > 0 \text{ such that for all } x \in \text{Dom}f \cap (a - \delta, a), \text{ we have } |f(x) - L_2| < \epsilon\)
    • Prove that \(\lim_{x \to a} f(x)\) exists if and only if \(\lim_{x \to a^+} f(x)\) and \(\lim_{x \to a^-} f(x)\) exist and are equal. Moreover, \(\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = \lim_{x \to a} f(x)\).

    Remark: If \(\lim_{x \to a^+} f(x) \neq \lim_{x \to a^-} f(x)\), then \(\lim_{x \to a} f(x)\) does not exist.

    Proof

    \(\Rightarrow \lim_{x \to a} f(x) = L \Rightarrow \lim_{x \to a^+} f(x)\) and \(\lim_{x \to a^-} f(x)\) exist and are equal.

    Exercise:

    \(\Rightarrow \lim_{x \to a^+} f(x)\), \(\lim_{x \to a^-} f(x)\) exist and \(\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(x) = L \Rightarrow \lim_{x \to a} f(x) = L\).

    \(\lim_{x \to a^+} f(x) = L\), if \(\forall \varepsilon > 0\), \(\exists \delta_1 > 0\) such that \(\forall x \in \text{Dom} f\) and \(x \in (a, a + \delta_1)\), we have \(|f(x) - L| < \varepsilon\).

    \(\lim_{x \to a^-} f(x) = L\), if \(\forall \varepsilon > 0\), \(\exists \delta_2 > 0\) such that \(\forall x \in \text{Dom} f\) and \(x \in (a - \delta_2, a)\), we have \(|f(x) - L| < \varepsilon\).

    If we choose \(\delta = \min\{\delta_1, \delta_2\}\), we have that \(|f(x) - L| < \varepsilon\).

  6. Analyze the existence of \(\lim_{x \to 0} \text{sgn}(x)\).

    Sign function: \(\text{Sgn}(x)=\begin{cases}1, & \text{if }x>0\\ 0, & \text{if }x=0\\ -1, & \text{if }x<0\end{cases}\)

    \(\lim_{x \to 0^+} \text{Sgn}(x) = \lim_{x \to 0^+} 1 = 1\)

    \(\lim_{x \to 0^-} \text{Sgn}(x) = \lim_{x \to 0^-} -1 = -1\)

    \(\lim_{x \to 0} \text{Sgn}(x)\) does not exist.

  7. Let \(g\) be the following function:

    \(g(x)=\begin{cases}2-x & \text{if }x<-1\\ x+2 & \text{if }-1\leq x<1\\ 4 & \text{if }x=1\\ 4-x & \text{if }x>1\end{cases}\)

    image​

    • Compute:

    • \(\lim_{x \to -1^+} g(x)\)

    • \(\lim_{x \to -1^-} g(x)\)
    • \(\lim_{x \to 1^+} g(x)\)
    • \(\lim_{x \to 1^-} g(x)\)

    \(\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} (x + 2) = 1 \quad (\neq \lim_{x \to 1^-} g(x))\)

    \(\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} (2 - x) = 3\)

    Since \(\lim_{x \to 1^+} g(x) \neq \lim_{x \to 1^-} g(x)\), \(\lim_{x \to 1} g(x)\) does not exist.

    \(\lim_{x \to 1^+} g(x) = 3\) and \(\lim_{x \to 1^-} g(x) = 3 \implies \lim_{x \to 1} g(x) = 3\).

  8. Limits at infinity:

    • \(\lim_{x \to \infty} f(x) = L \text{ if for all } \epsilon > 0, \text{ there exists } N > 0 \text{ such that whenever } x > N, \text{ it follows that } |f(x) - L| < \epsilon\)
    • \(\lim_{x \to -\infty} f(x) = L \text{ if for all } \epsilon > 0, \text{ there exists } N < 0 \text{ such that whenever } x < N, \text{ it follows that } |f(x) - L| < \epsilon\)

    • Prove that:

    • \(\lim_{x \to \infty} \frac{1}{x} = 0\)

    • \(\lim_{x \to -\infty} \frac{1}{x} = 0\)

  9. Analyze \(\lim_{x \to \infty} \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are real polynomials.