MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 5 August 2014

Lecture 12

Evaluate $lim_{x \to \infty} \frac{x^{2} - 7 x + 11}{3 x^{2} + 10} .$ $lim_{x \to \infty} \frac{x^{2} - 7 x + 11}{3 x^{2} + 10} = lim_{x \to \infty} \frac{1 - \frac{7}{x} + \frac{11}{x^{2}}}{3 + \frac{10}{x^{2}}} = \frac{1 - 0 + 0}{3 + 0} = \frac{1}{3} .$

Evaluate $lim_{x \to 0} \frac{sin 3 x}{sin 5 x} .$ $\begin{matrix} lim_{x \to 0} \frac{sin 3 x}{sin 5 x} & = & lim_{x \to 0} \frac{sin 3 x}{3 x} \cdot 3 x \frac{5 x}{sin 5 x} \frac{1}{5 x} \\ = & lim_{x \to 0} \frac{sin 3 x}{3 x} \frac{1}{\frac{sin 5 x}{5 x}} \frac{3 x}{5 x} \\ = & lim_{x \to 0} \frac{sin 3 x}{3 x} \frac{1}{\frac{sin 5 x}{5 x}} \cdot \frac{3}{5} \\ = & 1 \cdot \frac{1}{1} \cdot \frac{3}{5} = \frac{3}{5} . \end{matrix}$

Evaluate $lim_{x \to 1} \frac{1 - x}{{({cos}^{- 1} x)}^{2}} .$

Let $y = {cos}^{- 1} x .$ Then $y \to 0$ as $x \to 1$ and $x = cos y .$ So $\begin{matrix} lim_{x \to 1} \frac{1 - x}{{({cos}^{- 1} x)}^{2}} & = & lim_{y \to 0} \frac{1 - cos y}{y^{2}} \\ = & lim_{y \to 0} \frac{(1 - cos y)}{y^{2}} \frac{(1 + cos y)}{(1 + cos y)} \\ = & lim_{y \to 0} \frac{(1 - {cos}^{2} y)}{y^{2}} \frac{1}{1 + cos y} \\ = & lim_{y \to 0} \frac{sin y}{y} \frac{sin y}{y} \frac{1}{1 + cos y} \\ = & 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} . \end{matrix}$

$lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$ when $f (x) = sin 2 x .$ $\begin{matrix} lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x} & = & lim_{Δ x \to 0} \frac{sin (2 (x + Δ x)) - sin 2 x}{Δ x} \\ = & lim_{Δ x \to 0} \frac{sin (2 x + 2 Δ x) - sin 2 x}{Δ x} \\ = & lim_{Δ x \to 0} \frac{sin 2 x cos 2 Δ x + cos 2 x sin 2 Δ x - sin 2 x}{Δ x} \\ = & lim_{Δ x \to 0} sin 2 x \frac{(cos 2 Δ x - 1)}{Δ x} + cos 2 x \frac{sin 2 Δ x}{Δ x} \\ = & lim_{Δ x \to 0} sin 2 x \frac{cos 2 Δ x - 1}{2 Δ x} \cdot 2 + cos 2 x \frac{sin 2 Δ x}{2 Δ x} \cdot 2 \\ = & sin 2 x \cdot 0 \cdot 2 + cos 2 x \cdot 1 \cdot 2 = 2 cos 2 x . \end{matrix}$

$lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$ when $f (x) = cos x^{2} .$ $\begin{matrix} lim_{Δ x \to 0} \frac{cos {(x + Δ x)}^{2} - cos x^{2}}{Δ x} & = & lim_{Δ x \to 0} \frac{cos (x^{2} + 2 x Δ x + {(Δ x)}^{2}) - cos x^{2}}{Δ x} \\ = & lim_{Δ x \to 0} \frac{cos x^{2} cos (2 x Δ x + {(Δ x)}^{2}) - sin x^{2} sin (2 x Δ x + {(Δ x)}^{2}) - cos x^{2}}{Δ x} \\ = & lim_{Δ x \to 0} cos x^{2} \frac{(cos (2 x Δ x + {(Δ x)}^{2} - 1))}{Δ x} - sin x^{2} \frac{sin (2 x Δ x + {(Δ x)}^{2})}{Δ x} \\ = & lim_{Δ x \to 0} cos x^{2} \frac{(cos (2 x Δ x + {(Δ x)}^{2}) - 1)}{2 x Δ x + {(Δ x)}^{2}} \frac{(2 x Δ x + {(Δ x)}^{2})}{Δ x} \\ - sin x^{2} \frac{(sin (2 x Δ x + {(Δ x)}^{2}))}{2 x Δ x + {(Δ x)}^{2}} \frac{(2 x Δ x + {(Δ x)}^{2})}{Δ x} \\ = & lim_{Δ x \to 0} cos x^{2} \frac{(cos (2 x Δ x + {(Δ x)}^{2}) - 1)}{2 x Δ x + {(Δ x)}^{2}} \cdot (2 x + Δ x) - sin x^{2} \frac{(sin (2 x Δ x + {(Δ x)}^{2}))}{2 x Δ x + {(Δ x)}^{2}} \cdot (2 x + Δ x) \\ = & cos x^{2} \cdot 0 \cdot 2 x - sin x^{2} \cdot 1 \cdot 2 x = - 2 x sin x^{2} . \end{matrix}$

$lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x}$ when $f (x) = x^{x} .$ $\begin{matrix} lim_{Δ x \to 0} \frac{{(x + Δ x)}^{x + Δ x} - x^{x}}{Δ x} & = & lim_{Δ x \to 0} \frac{{(e^{ln (x + Δ x)})}^{x + Δ x} - {(e^{ln x})}^{x}}{Δ x} \\ = & lim_{Δ x \to 0} \frac{e^{(x + Δ x) ln (x + Δ x)} - e^{x ln x}}{Δ x} \\ = & lim_{Δ x \to 0} e^{x ln x} \frac{(e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1)}{Δ x} \\ = & lim_{Δ x \to 0} e^{x ln x} \frac{(e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1)}{((x + Δ x) ln (x + Δ x) - x ln x)} \frac{((x + Δ x) ln (x + Δ x) - x ln x)}{Δ x} \\ = & lim_{Δ x \to 0} e^{x ln x} (\frac{e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1}{(x + Δ x) ln (x + Δ x) - x ln x}) (\frac{x ln (x + Δ x) - x ln x}{Δ x} + ln (x + Δ x)) \\ = & lim_{Δ x \to 0} e^{x ln x} (\frac{e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1}{(x + Δ x) ln (x + Δ x) - x ln x}) (\frac{x ln (x (1 + \frac{Δ x}{x})) - x ln x}{Δ x} + ln (x + Δ x)) \\ = & lim_{Δ x \to 0} e^{x ln x} (\frac{e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1}{(x + Δ x) ln (x + Δ x) - x ln x}) (\frac{x (ln x + ln (1 + \frac{Δ x}{x})) - x ln x}{Δ x} + ln (x + Δ x)) \\ = & lim_{Δ x \to 0} e^{x ln x} (\frac{e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1}{(x + Δ x) ln (x + Δ x) - x ln x}) (\frac{x ln (1 + \frac{Δ x}{x})}{Δ x} + ln (x + Δ x)) \\ = & lim_{Δ x \to 0} e^{x ln x} (\frac{e^{(x + Δ x) ln (x + Δ x) - x ln x} - 1}{(x + Δ x) ln (x + Δ x) - x ln x}) (\frac{ln (1 + \frac{Δ x}{x})}{\frac{Δ x}{x}} + ln (x + Δ x)) \\ = & e^{x ln x} \cdot 1 (1 + ln x) \\ = & x^{x} + x^{x} ln x . \end{matrix}$

Notes and References

These are a typed copy of Lecture 12 from a series of handwritten lecture notes for the class MATH 221 given on October 4, 2000.

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