Infi-1 24

Infi-1 24

Darboux's theorem #theorem

\begin{matrix} Let f be a differentiable function on [a, b] \\ Then \forall a_{1} < b_{1} \in [a, b] : \forall y \in [f^{'} (a_{1}), f^{'} (b_{1})] \exists x \in [a, b] : f^{'} (x) = y \\ In other words, f^{'} has intermediate value property \\ Note: f^{'} does not have to be differentiable on [a, b] \\ If it is not continuous, then the discontinuities will be essential \end{matrix}

Example

\begin{matrix} f (x) = {\begin{cases} x^{2} \sin (\frac{1}{x^{7}}) & x \neq 0 \\ 0 & x = 0 \end{cases} \\ f^{'} (0) = lim_{h \to 0} \frac{f (h) - f (0)}{h} = lim_{h \to 0} \frac{h^{2} \sin (\frac{1}{x^{7}})}{h} = 0 \\ f^{'} (x) = 2 x \sin (\frac{1}{x^{7}}) + x^{2} \cos (\frac{1}{x^{7}}) \cdot \frac{- 7}{x^{8}} = 2 x \sin (\frac{1}{x^{7}}) - \frac{7 \cos (\frac{1}{x^{7}})}{x^{6}} \\ f is continuous at 0 \\ f^{'} is not continuous (essential discontinuity) at 0 \end{matrix}

L'Hopital's theorem/rule #theorem

\begin{matrix} Let f, g be differentiable functions in the locality of a \\ Let \exists lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} \\ Let lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0 or \pm \infty \\ Then lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} \end{matrix}

Example

\begin{matrix} lim_{x \to 0} \frac{\overset{\to 0}{\overset{⏞}{\arctan x}}}{\underset{\to 0}{\underset{⏟}{e^{x} - 1}}} \\ (\arctan x)^{'} = \frac{1}{1 + x^{2}} \\ (e^{x} - 1)^{'} = e^{x} \\ By L’Hopital’s rule: lim_{x \to 0} \frac{\arctan x}{e^{x} - 1} = lim_{x \to 0} \frac{1}{(1 + x^{2}) e^{x}} = 1 \end{matrix}

Example

\begin{matrix} lim_{x \to 0} \frac{x - \arctan x}{x - \sin x} \\ By L’Hopital’s rule: lim_{x \to 0} \frac{(1 - \frac{1}{1 + x^{2}})}{1 - \cos x} = lim_{x \to 0} \frac{x^{2}}{(1 + x^{2}) (1 - \cos x)} = \\ = lim_{x \to 0} \frac{1}{1 + x^{2}} \cdot lim_{x \to 0} \frac{x^{2}}{1 - \cos x} = lim_{x \to 0} \frac{x^{2}}{1 - \cos x} \\ lim_{x \to 0} \frac{1 - \cos x}{x^{2}} = \frac{1}{2} ⟹ lim_{x \to 0} \frac{x^{2}}{1 - \cos x} = 2 \\ ⟹ lim_{x \to 0} \frac{x - \arctan x}{x - \sin x} = 2 \end{matrix}

Example

\begin{matrix} lim_{x \to \infty} \frac{e^{x}}{x^{3} + 2} = lim_{x \to \infty} \frac{e^{x}}{3 x^{2}} = lim_{x \to \infty} \frac{e^{x}}{6 x} = lim_{n \to \infty} \frac{e^{x}}{6} = \infty \end{matrix}

Example

\begin{matrix} lim_{x \to 7^{+}} (x - 7) \ln (x - 7) = lim_{x \to 7^{+}} \frac{\ln (x - 7)}{\frac{1}{x - 7}} = lim_{x \to 7^{+}} \frac{x - 7}{\frac{1}{\ln (x - 7)}} \\ lim_{x \to 7^{+}} \frac{\ln (x - 7)}{\frac{1}{x - 7}} = lim_{x \to 7^{+}} \frac{\frac{1}{x - 7}}{- \frac{1}{(x - 7)^{2}}} = lim_{x \to 7^{+}} - (x - 7) = 0 \end{matrix}

Example

\begin{matrix} lim_{x \to 0^{+}} (\sin x)^{\sin x} = lim_{x \to 0} e^{\sin x \ln (\sin x)} = lim_{x \to 0} e^{\ln (\sin x) / (\sin x)^{- 1}} \\ lim_{x \to 0^{+}} \frac{\ln (\sin x)}{(\sin x)^{- 1}} = \frac{\frac{\cos x}{\sin x}}{- (\sin x)^{- 2} \cdot \cos x} = lim_{x \to 0^{+}} - \frac{(\sin x)^{2}}{\sin x} = 0 \\ ⟹ lim_{x \to 0} (\sin x)^{\sin x} = e^{0} = 1 \end{matrix}

Example

\begin{matrix} lim_{x \to \infty} {(\sin (\frac{1}{x}) + \cos (\frac{1}{x}))}^{x^{2}} = lim_{x \to \infty} e^{x^{2} \ln (\sin (1 / x) + \cos (1 / x))} \\ lim_{x \to \infty} \frac{\ln (\sin (\frac{1}{x}) + \cos (\frac{1}{x}))}{\frac{1}{x^{2}}} \underset{t = \frac{1}{x}}{\underset{⏟}{=}} lim_{t \to 0^{+}} \frac{\ln (\sin t + \cos t)}{t^{2}} \overset{L}{=} lim_{t \to 0^{+}} \frac{\frac{\cos t - \sin t}{\sin t + \cos t}}{2 t} = \infty \end{matrix}

When doesn't L'Hopital's rule work? #lemma

Case 1

\begin{matrix} Exist functions, such that their differentiation only makes things worse \\ lim_{x \to 0} \frac{e^{- 1 / x^{2}}}{x^{2}} \overset{L}{=} lim_{x \to 0} \frac{\frac{2}{x^{3}} e^{- 1 / x^{2}}}{2 x} = \frac{e^{- 1 / x^{2}}}{x^{4}} \end{matrix}

Case 2

\begin{matrix} lim_{x \to a} \frac{f^{'}}{g^{'}} doesn’t exist \\ lim_{x \to \infty} \frac{x + \sin x}{x + \cos x} \overset{L}{=} lim_{x \to \infty} \frac{1 + \cos x}{1 - \sin x} - doesn’t exist \\ But lim_{x \to \infty} \frac{1 + \frac{\sin x}{x}}{1 + \frac{\cos x}{x}} = 1 - exists \end{matrix}

Case 3

\begin{matrix} lim_{x \to 0} \frac{\sin (x)}{x} \overset{L}{=} lim_{x \to 0} \frac{\cos x}{1} = 1 \\ Proof that (\sin x)^{'} = \cos x uses the fact that lim_{x \to 0} \frac{\sin x}{x} = 1 \\ Using L’Hopital’s rule here would create a cyclic dependency \\ lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \\ lim_{x \to 0} \frac{\ln (1 + x)}{x} = 1 \\ lim_{x \to 0} \frac{a^{x} - 1}{x} = \ln a \end{matrix}

Case 4

\begin{matrix} Special limits \\ \frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, 1^{\infty}, 0^{0}, \infty^{0} \\ With these 6 forms we can use L’Hopital’s rule \\ But what do we do with \infty - \infty ? \\ In this case there is no universal technique to convert this into a \frac{0}{0} or \frac{\infty}{\infty} division \\ In some cases we will be able to find it, in some cases not \end{matrix}

Logarithm vs polynomial #lemma

\begin{matrix} \forall a, b > 0 : lim_{x \to \infty} \frac{(\ln x)^{a}}{x^{b}} = 0 \\ Proof: \\ lim_{x \to \infty} \frac{(\ln x)^{a}}{x^{b}} = lim_{x \to \infty} {(\frac{\ln (x)}{x^{b / a}})}^{a} \\ t \to 0 ⟹ t^{a} \to 0^{a} = 0 \\ lim_{x \to \infty} \frac{\ln x}{x^{b / a}} \overset{L}{=} lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{b}{a} \cdot x^{\frac{b}{a} - 1}} = lim_{x \to \infty} \frac{1}{\frac{b}{a} \cdot x^{b / a}} = 0 \\ ⟹ lim_{x \to \infty} \frac{(\ln x)^{a}}{x^{b}} = 0 \end{matrix}

Polynomial vs exponential #lemma

\begin{matrix} \forall a > 1, b > 0 : lim_{x \to \infty} \frac{x^{b}}{a^{x}} = 0 \\ Proof: \\ lim_{x \to \infty} \frac{x^{b}}{a^{x}} = lim_{x \to \infty} {(\frac{x}{a^{x / b}})}^{b} \\ lim_{x \to \infty} \frac{x}{a^{x / b}} \overset{L}{=} lim_{x \to \infty} \frac{1}{\ln (a) \cdot a^{x / b} \cdot \frac{1}{b}} = 0 \\ ⟹ lim_{x \to \infty} \frac{x^{b}}{a^{x}} = 0 \end{matrix}

\begin{matrix} lim_{x \to 0} \frac{a^{x} - 1}{x} = \ln a \\ \sum \frac{\ln (n)}{n} \geq \sum \frac{1}{n} ⟹ \sum \frac{\ln (n)}{n} diverges \end{matrix}

\begin{matrix} lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} \\ Let x_{n} = n \\ Let f (x_{n}) = (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} \\ Then lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} = lim_{x \to \infty} f (x) \\ lim_{x \to \infty} (e^{1 / x} - e^{- 1 / x})^{\sin (1 / x)} = lim_{x \to \infty} e^{\sin (1 / x) \cdot \ln (e^{1 / x} - e^{- 1 / x})} \\ lim_{x \to \infty} \sin (1 / x) \cdot \ln (e^{1 / x} - e^{- 1 / x}) = lim_{x \to \infty} \frac{\ln (e^{1 / x} - e^{- 1 / x})}{{(\sin (\frac{1}{x}))}^{- 1}} \\ Let t = \frac{1}{x}, t \to 0^{+} \\ lim_{t \to 0} \frac{\ln (e^{t} - e^{- t})}{(\sin t)^{- 1}} \overset{L}{=} lim_{t \to 0} \frac{\frac{e^{t} + e^{- t}}{e^{t} - e^{- t}}}{- (\sin t)^{- 2} \cdot \cos t} = - lim_{t \to 0} \frac{(e^{t} + e^{- t}) (\sin t)^{2}}{(e^{t} - e^{- t}) \cos t} = \\ \overset{L}{=} - lim_{t \to 0} \frac{e^{t} + e^{- t}}{\cos t} \cdot lim_{t \to 0} \frac{(\sin t)^{2}}{e^{t} - e^{- t}} = - 2 \cdot lim_{t \to 0} \frac{2 \overset{\to 0}{\overset{⏞}{\sin t}} \cdot \overset{\to 1}{\overset{⏞}{\cos t}}}{\underset{\to 2}{\underset{⏟}{e^{t} + e^{- t}}}} = - 2 \cdot 0 = 0 \\ ⟹ lim_{x \to \infty} e^{\sin (1 / x) \cdot \ln (e^{1 / x} - e^{- 1 / x})} = e^{0} = 1 \\ ⟹ lim_{n \to \infty} (e^{1 / n} - e^{- 1 / n})^{\sin (1 / n)} = 1 \end{matrix}