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11.28 Infinite limits

Definition

\(\lim_{x \to x_0} f(x) = +\infty\) if \(\forall M > 0\), \(\exists \delta > 0\): whenever \(0 < |x-x_0| < \delta\), it follows that \(f(x) > M\)

\(\lim_{x \to x_0} f(x) = -\infty\) if \(\forall N < 0\), \(\exists \delta > 0\): whenever \(0 < |x-x_0| < \delta\), it follows that \(f(x) < N\)

Example

\(f(x) = \frac{1}{|x|}\), Dom \(f = \mathbb{R} - \{0\}\)

\(\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = 0\)

Prove that \(\lim_{x \to 0} f(x) = +\infty\)

We want \(0<|x|<\delta\Rightarrow \frac{1}{|x|}>M\), then we can just take \(\delta=\frac{1}{M}\)​

Definition

\(\lim_{x \to x_0^+} f(x) = +\infty\) if \(\forall M > 0\), \(\exists \delta > 0\): \(\forall x \in (x_0, x_0+\delta)\), \(f(x) > M\)
\(\lim_{x \to x_0^-} f(x) = -\infty\) if \(\forall N < 0\), \(\exists \delta > 0\): \(\forall x \in (x_0-\delta, x_0)\), \(f(x) < N\)
(2 more similar)

Example

\(f(x) = \frac{1}{x}\), prove \(\lim_{x \to 0^+} f(x) = +\infty\) and \(\lim_{x \to 0^-} f(x) = -\infty\)

Proof
Let \(M > 0\) be arbitrary, take \(\delta < \frac{1}{M}\), \(x \in (0, 0+\delta)\) \(\Rightarrow |x-0| < \frac{1}{M}\) \(\Rightarrow x < \frac{1}{M}\)
\(\Rightarrow \frac{1}{x} > M\) \(\Rightarrow \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1}{x} = +\infty\)

Let \(N < 0\) be arbitrary, take \(\delta < -\frac{1}{N}\), \(x \in (0-\delta, 0)\) \(\Rightarrow |x-0| < -\frac{1}{N}\) \(\Rightarrow x > \frac{1}{N}\)
\(\Rightarrow \frac{1}{x} < N\) \(\Rightarrow \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1}{x} = -\infty\)

Definition

The vertical line \(x = a\) is a vertical asymptote of \(f\) if at least one of the following hold: \(\lim_{x \to a} f(x) = \pm\infty\), \(\lim_{x \to a^+} f(x) = \pm\infty\), \(\lim_{x \to a^-} f(x) = \pm\infty\)

The horizontal line \(y = L\) is a horizontal asymptote of \(f\) if \(\lim_{x \to \infty} f(x) = L\) or \(\lim_{x \to -\infty} f(x) = L\)

Example

Prove that \(\lim_{x \to a} f(x) = 0\) iff \(\lim_{x \to a} |f(x)| = 0\)

Proof

\(\Rightarrow\)) \(\forall \varepsilon > 0\), \(\exists \delta > 0\): \(\forall x \in\) Dom \(f\) and \(0 < |x-a| < \delta\), \(|f(x)-0| < \varepsilon\)
Then \(|f(x) - 0| = |f(x)| = |f|(x) = ||f|(x)-0| < \varepsilon\)
Thus For the same \(\delta\), \(|f|\) satisfies the condition \(\Rightarrow \lim_{x \to a} |f(x)| = 0\)

\(\Leftarrow\)) \(\forall \varepsilon > 0\), \(\exists \delta > 0\): \(\forall x \in\) Dom \(f\) and \(0 < |x-a| < \delta\), \(||f|(x)-0| < \varepsilon\)
Then \(||f(x)| - 0| = ||f(x)|| = |f(x)| = |f(x)-0| < \varepsilon\)
Thus For the same \(\delta\), \(f\) satisfies the condition \(\Rightarrow \lim_{x \to a} f(x) = 0\)

Example

Prove that \(\lim_{x \to a} f(x) = L \Rightarrow \lim_{x \to a} |f(x)| = |L|\)

Proof

\(\forall \varepsilon > 0\), \(\exists \delta > 0\): \(\forall x \in\) Dom \(f\) and \(0 < |x-a| < \delta\), \(|f(x)-L| < \varepsilon\)
By triangular inequality, \(||f(x)| - |L|| < |f(x)-L| < \varepsilon \Rightarrow ||f(x)-|L|| < \varepsilon\)
\(\Rightarrow \lim_{x \to a} |f(x)| = |L|\) with the same \(\delta\)

The converse does not always hold (e.g. \(y = \text{sign}(x)\))

Counterexample

\(\text{Sign}(x) = \begin{cases} 1 & x > 0 \\ 0 & x = 0 \\ -1 & x < 0 \end{cases}\) image​

Since \(\lim_{x \to 0^-} \text{Sign}(x) = -1\) and \(\lim_{x \to 0^+} \text{Sign}(x) = 1\), then \(\lim_{x \to 0} \text{Sign}(x) \text{ does not exist}\)

But \(\lim_{x \to 0} |\text{Sign}(x)| = \lim_{x \to 0} 1 = 1\)

Example

  1. \(|\sin x| \leq |x|\), \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)

    Proof

    Let \(C = \{(a,b): a^2+b^2 = 1\}\) Unit circle image
    ​\(b = \sin x\), \(a = \cos x\)

    Consider the arc from \((1,0)\) to \((a,b)\), length of the arc is \(s\) image​

    Then \(s=r\theta=1\cdot x=x\), then we compare the area image \(A_1=\frac{\sin x}{2},A_2=\frac{x}{2}r^2=\frac{x}{2}\)
    Since \(A_1<A_2\), then \(\sin x<x\), \(\forall x\in[0,\frac{\pi}{2}]\)

    Then \(\sin (-x)>-x\), \(\forall x\in[-\frac{\pi}{2},0]\), then \(|\sin x|\leq|x|,\forall-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}\)​

  2. \(\lim_{x \to 0} \sin x = 0\)

    Since \(-x \leq \sin x \leq x\), then \(\lim_{x \to 0} x = \lim_{x \to 0} (-x) = 0 \Rightarrow \lim_{x \to 0} \sin x = 0\)

  3. \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)

    image​

    Proof: Consider the areas of triangles image
    ​\(A_1 = \frac{\sin x}{2}\) \(A_2 = \frac{x}{2}\) \(A_3 = \frac{\tan x}{2}\)
    ​\(A_1 \leq A_2 \leq A_3\)
    ​\(\frac{2}{\sin x} \frac{\sin x}{2} \leq \frac{2}{\sin x} \frac{x}{2} \leq \frac{2}{\sin x} \frac{\tan x}{2}\)
    ​\(1 \leq \frac{x}{\sin x} \leq \frac{1}{\cos x}\)
    ​\(1 \geq \frac{\sin x}{x} \geq \cos x\)
    ​\(\lim_{x \to 0} 1 = \lim_{x \to 0} \cos x = 1 \Rightarrow \lim_{x \to 0} \frac{\sin x}{x} = 1\)

  4. \(\lim_{x\to0}\frac{\sin3x}{x}=\lim_{x\to0}3\cdot\frac{\sin3x}{3x}=3\) \((t=3x)\)

  5. \(\lim_{x \to 0} \frac{1-\cos x}{x} = \lim_{x \to 0} \frac{(1-\cos x)(1+\cos x)}{x(1+\cos x)} = \lim_{x \to 0} \frac{\sin^2 x}{x(1+\cos x)}\)
    ​\(= \lim_{x \to 0} \frac{\sin x}{1+\cos x} = \lim_{x \to 0} \tan \frac{x}{2} = 0\)

  6. \(\lim_{x \to \infty} x\sin \frac{1}{x} = \lim_{x \to \infty} \frac{\sin(\frac{1}{x})}{\frac{1}{x}}\) \((l = \frac{1}{x})\) \(= \lim_{l \to 0} \frac{\sin l}{l} = 1\)