1 Solution 1
\(\lim_{x \to \dfrac{1}{2}} tan (\pi x^2)\)
By the continuity of the \(tan\) function,
\(\lim_{x \to \dfrac{1}{2}} tan (\pi x^2) = tan \dfrac{\pi}{4} = 1\)
2 Solution 2
\(\lim_{x \to 6} \dfrac{\sqrt{x+3} - 2}{x-1}\)
Since \(x - 1 \neq 0\), \(\dfrac{\sqrt{x+3} - 2}{x-1}\) is continous at \(6\).
So, \(\lim_{x \to 6} \dfrac{\sqrt{x+3} - 2}{x-1}\)
\(= \dfrac{\sqrt{9} - 2}{5}\)
\(= \dfrac{3-2}{5} = \dfrac{1}{5}\)
3 Solution 3
\(\lim_{x \to 1} \dfrac{\sqrt{x+3} - 2}{x-1}\)
Both the numerator and denominator approach \(0\) as \(x\) approaches \(1\).
\(\lim_{x \to 1} \dfrac{\sqrt{x+3} - 2}{x-1} * \dfrac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2}\)
\(= \lim_{x \to 1} \dfrac{x + 3 - 4}{(x-1)(\sqrt{x+3} + 2)}\)
\(= \lim_{x \to 1} \dfrac{1}{\sqrt{x+3} + 2}\)
Since the rational function is continous,
\(= \dfrac{1}{\sqrt{4} + 2} = \dfrac{1}{4}\)
4 Solution 4
\(\lim_{x \to 0} \dfrac{\sqrt{4+x} - \sqrt{4-x}}{x}\)
Both the numerator and denominator approach \(0\) as \(x\) approaches \(0\).
\(\lim_{x \to 0} \dfrac{\sqrt{4+x} - \sqrt{4-x}}{x} * \dfrac{\sqrt{4+x} + \sqrt{4-x}}{\sqrt{4+x} + \sqrt{4-x}}\)
\(= \lim_{x \to 0} \dfrac{2}{\sqrt{4+x} + \sqrt{4-x}}\)
Since the rational function is continous,
\(\dfrac{2}{\sqrt{4} + \sqrt{4}} = \dfrac{2}{2+2} = \dfrac{2}{4} = \dfrac{1}{2}\)
5 Solution 5
\(\lim_{t \to 1} \dfrac{\sqrt{t^2+3t} - 2}{t-1}\)
Both the numerator and denominator approach \(0\) as \(t\) approaches \(1\).
\(\lim_{t \to 1} \dfrac{\sqrt{t^2+3t} - 2}{t-1} * \dfrac{\sqrt{t^2+3t} + 2}{\sqrt{t^2+3t} + 2}\)
\(= \lim_{t \to 1} \dfrac{(t-1)(t+4)}{(t-1)(\sqrt{t^2 + 3t} + 2)}\)
\(= \lim_{t \to 1} \dfrac{t+4}{\sqrt{t^2 + 3t} + 2}\)
Since the above rational function is continous,
\(\dfrac{1+4}{\sqrt{1+3} + 2} = \dfrac{5}{2+2} = \dfrac{5}{4}\)
6 Solution 6
\(\lim_{x \to 4} \dfrac{x-2-\sqrt{x}}{x-4}\)
Both the numerator and denominator approach \(0\) as \(x\) approaches \(4\).
\(\lim_{x \to 4} \dfrac{x-2-\sqrt{x}}{x-4} * \dfrac{x-2+\sqrt{x}}{x-2+\sqrt{x}}\)
\(= \lim_{x \to 4} \dfrac{x-1}{(x-2) + \sqrt{x}}\)
Since the above rational function is continous,
\(= \dfrac{3}{x+ \sqrt{4}} = \dfrac{3}{4}\)
7 Solution 7
\(\lim_{z \to 1^+} \dfrac{z-1}{\sqrt{z^2 - 1}}\)
Both the numberator and denominator approach \(0\) as z approaches \(1^+\).
\(\lim_{z \to 1^+} \dfrac{z-1}{\sqrt{z - 1} \sqrt{z+1}}\)
\(= \lim_{z \to 1^+} \dfrac{\sqrt{z-1}}{\sqrt{z+1}}\)
Since the above rational function is continous,
\(= \dfrac{0}{\sqrt{2}} = 0\)
8 Solution 8
\(\lim_{x \to 4} \dfrac{1}{\sqrt{x} - 2} - \dfrac{4}{x-4}\)
Let's simplify the function,
\(\dfrac{1}{\sqrt{x} - 2} - \dfrac{4}{x-4}\)
\(= \dfrac{x-4-4(\sqrt{x} - 2)}{(x-4)(\sqrt{x}-2)}\)
\(= \dfrac{x-4-4\sqrt{x}+8}{(x-4)(\sqrt{x}-2)}\)
\(= \dfrac{x+4-4\sqrt{x}}{(x-4)(\sqrt{x}-2)}\)
\(= \dfrac{x+4-4\sqrt{x}}{(x-4)(\sqrt{x}-2)} * \dfrac{x+4+4\sqrt{x}}{x+4+4\sqrt{x}}\)
\(= \dfrac{(x-4)^2}{(x-4)(\sqrt{x}-2)(x+4+4\sqrt{x})}\)
\(= \dfrac{x-4}{(\sqrt{x}-2)(x+4+4\sqrt{x})}\)
\(= \dfrac{x-4}{(\sqrt{x}-2)(x+4+4\sqrt{x})} * \dfrac{\sqrt{x} + 2}{\sqrt{x} + 2}\)
\(= \dfrac{\sqrt{x} + 2}{x+4+4\sqrt{x}}\)
Since the above rational function is continous,
\(= \dfrac{2+2}{4+4+4.2} = \dfrac{4}{4(1+1+2)} = \dfrac{1}{4}\)
9 Solution 9
\(\lim_{x \to 8} \dfrac{\sqrt[3]{x}-2}{x-8}\)
Both the numberator and denominator approach \(0\) as x approaches \(8\).
\(x-8 = (\sqrt[3]{x})^3 - 2^3\) \(= (\sqrt[3]{x} - 2)(x^{\dfrac{2}{3}} + 2.\sqrt[3]{x} + 4)\)
Simplifying the above limits, we get
\(= \lim_{x \to 8} \dfrac{1}{x^{\dfrac{2}{3}} + + 2.\sqrt[3]{x} + 4}\)
Since the above rational function is continous,
\(= \dfrac{1}{2^2 + 2.2 + 4} = \dfrac{1}{4+4+4} = \dfrac{1}{12}\)
10 Solution 10
\(\lim_{x \to 0^+} \dfrac{1}{\sqrt[3]{x}} - \dfrac{1}{\sqrt{x}}\)
Let's simplify the function,
\(= x^{-\dfrac{1}{3}} - x^{-\dfrac{1}{2}}\)
\(= x^{-\dfrac{1}{2}}(\dfrac{x^{-\dfrac{1}{3}}}{x^{-\dfrac{1}{2}}} - 1)\)
\(= x^{-\dfrac{1}{2}}(x^{-\dfrac{1}{6}} - 1)\)
Let \(f(x) = \sqrt{x}\)
\(\lim_{x \to 0^+} f(x) = 0\)
\(\lim_{x \to 0^+} \dfrac{1}{f(x)} = \infty\)
\(\lim_{x \to 0^+} x^{\dfrac{1}{6}} - 1 = -1\)
From theorems 2.5.4,
\(\lim_{x \to 0^+} \dfrac{1}{\sqrt[3]{x}} - \dfrac{1}{\sqrt{x}} = -\infty\)
11 Solution 11
\(\lim_{u \to \infty} \dfrac{\sqrt{u^2 + 9}}{3u}\)
\(= \lim_{u \to \infty} \dfrac{\sqrt{u^2(1+\dfrac{9}{u^2})}}{3u}\)
\(= \lim_{u \to \infty} \dfrac{\sqrt{(1+\dfrac{9}{u^2})}}{u}\)
\(= \dfrac{\sqrt{1 + 0}}{3}\)
\(= \dfrac{1}{3}\)
12 Solution 12
\(\lim_{x \to \infty} \sqrt{x^2 + 100} - x\)
\(= \lim_{x \to \infty} \sqrt{x^2 + 100} - x * \dfrac{\sqrt{x^2 + 100} + x}{\sqrt{x^2 + 100} + x}\)
\(= \lim_{x \to \infty} \dfrac{x^2 + 100 - x^2}{\sqrt{x^2 + 100} + x}\)
\(= \lim_{x \to \infty} \dfrac{100}{x(\sqrt{1 + \dfrac{100}{x^2}} + 1)}\)
\(= 0\)
13 Solution 13
\(\lim_{x \to -\infty} \sqrt{x^2 + x} + x\)
\(= \lim_{x \to -\infty} \sqrt{x^2 + x} + x * \dfrac{\sqrt{x^2 + x} - x}{\sqrt{x^2 + x} - x}\)
\(= \lim_{x \to -\infty} \dfrac{x^2 + x - x^2}{\sqrt{x^2}(\sqrt{1 + \dfrac{1}{x}}) - x}\)
\(= \lim_{x \to -\infty} \dfrac{x}{-x(\sqrt{1 + \dfrac{1}{x}}) - x}\)
\(= \lim_{x \to -\infty} \dfrac{-1}{(\sqrt{1 + \dfrac{1}{x}}) + 1}\)
\(= \dfrac{-1}{\sqrt{1} + 1} = \dfrac{-1}{2}\)
14 Solution 14
\(\lim_{x \to \infty} \sqrt{x^4 + x} - x^2\)
\(= \lim_{x \to \infty} \sqrt{x^4 + x} - x^2 * \dfrac{\sqrt{x^4 + x} + x^2}{\sqrt{x^4 + x} + x^2}\)
\(= \lim_{x \to \infty} \dfrac{x^4 + x - x^4}{\sqrt{x^4 + x} + x^2}\)
\(= \lim_{x \to \infty} \dfrac{x}{x^2(\sqrt{1 + \dfrac{1}{x^3}} + 1)}\)
\(= \lim_{x \to \infty} \dfrac{1}{x(\sqrt{1 + \dfrac{1}{x^3}} + 1)}\)
\(= 0\)
15 Solution 15
\(\lim_{x \to \infty} \sqrt{x^4 + x^2} - x^2\)
\(= \lim_{x \to \infty} \sqrt{x^4 + x^2} - x^2 * \dfrac{\sqrt{x^4 + x^2} + x^2}{\sqrt{x^4 + x^2} + x^2}\)
\(= \lim_{x \to \infty} \dfrac{x^4 + x^2 - x^4}{\sqrt{x^4 + x^2} + x^2}\)
\(= \lim_{x \to \infty} \dfrac{x^2}{x^2(\sqrt{1 + \dfrac{1}{x^2}} + 1)}\)
\(= \lim_{x \to \infty} \dfrac{1}{\sqrt{1 + \dfrac{1}{x^2}} + 1} = \dfrac{1}{\sqrt{1} + 1} = \dfrac{1}{2}\)
16 Solution 16
\(\lim_{x \to \infty} \sqrt{x^4 + x^3} - x^2\)
\(= \lim_{x \to \infty} \sqrt{x^4 + x^3} - x^2 * \dfrac{\sqrt{x^4 + x^3} + x^2}{\sqrt{x^4 + x^3} + x^2}\)
\(= \lim{x \to \infty} \dfrac{x^4 + x^3 - x^4}{\sqrt{x^4 + x^3} + x^2}\)
\(= \lim{x \to \infty} \dfrac{x^3}{x^2(\sqrt{1 + \dfrac{1}{x}} + 1)}\)
\(= \infty\)
17 Solution 17
\(\lim_{\theta \to \dfrac{\pi}{2}} \dfrac{sin 2\theta}{cos \theta}\)
\(= \lim_{\theta \to \dfrac{\pi}{2}} \dfrac{2sin\theta cos\theta}{cos\theta}\)
\(= \lim_{\theta \to \dfrac{\pi}{2}} 2 sin \theta\)
\(= 2 sin \dfrac{\pi}{2} = 2\)
18 Solution 18
\(\lim_{\theta \to \dfrac{\pi}{4}} \dfrac{\sqrt{2} - 2cos\theta}{1 - 2cos^2 \theta}\)
\(= \lim_{\theta \to \dfrac{\pi}{4}} \dfrac{\sqrt{2} - 2cos\theta}{1 - 2cos^2 \theta} * \dfrac{\sqrt{2} + 2cos\theta}{\sqrt{2} + 2cos\theta}\)
\(= \lim_{\theta \to \dfrac{\pi}{4}} \dfrac{2-4cos^2\theta}{(sin^2\theta - cos^2 \theta)(\sqrt{2} + 2cos\theta)}\)
\(= \lim_{\theta \to \dfrac{\pi}{4}} \dfrac{2(sin^2\theta - cos^2 \theta)}{(sin^2\theta - cos^2 \theta)(\sqrt{2} + 2cos\theta)}\)
\(= \lim_{\theta \to \dfrac{\pi}{4}} \dfrac{2}{\sqrt{2} + 2cos\theta}\)
\(= \lim_{\theta \to \dfrac{\pi}{4}} \dfrac{2}{\sqrt{2} + 2\dfrac{1}{\sqrt{2}}}\)
\(= \dfrac{2.\sqrt{2}}{2+2} = \dfrac{1}{\sqrt{2}}\)
19 Solution 19
\(\lim_{\theta \to \dfrac{\pi}{4}} \dfrac{cos\theta - sin\theta}{1-2sin^2 \theta}\)
Since \(sin^2 \theta + cos^2 \theta = 1\),
\(\lim_{\theta \to \dfrac{\pi}{4}} \dfrac{cos \theta - sin \theta}{cos^2 \theta - sin^2 \theta}\)
\(\lim_{\theta \to \dfrac{\pi}{4}} \dfrac{cos \theta - sin \theta}{(cos \theta - sin \theta)(cos \theta + sin \theta)}\)
\(\lim_{\theta \to \dfrac{\pi}{4}} \dfrac{1}{cos \theta + sin \theta}\)
\(= \dfrac{1}{cos \dfrac{\pi}{4} + sin \dfrac{\pi}{4}}\)
\(= \dfrac{1}{\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}}\)
\(= \dfrac{1}{\sqrt{2}}\)
20 Solution 20
\(\lim_{\theta \to 0} \dfrac{1 - cos\theta}{\theta}\)
\(= \lim_{\theta \to 0} \dfrac{1 - cos\theta}{\theta} * \dfrac{1 + cos\theta}{1 + cos\theta}\)
\(= \lim_{\theta \to 0} \dfrac{1-cos^2 \theta}{\theta(1 + cos \theta)}\)
\(= \lim_{\theta \to 0} \dfrac{sin^2 \theta}{\theta(1 + cos \theta)}\)
\(= \lim_{\theta \to 0} sin \theta \dfrac{sin \theta}{\theta} \dfrac{1}{1 + cos \theta}\)
\(= 0.1.\dfrac{1}{2} = 0\)
21 Solution 21
\(\lim_{\theta \to 0} \dfrac{1 - cos\theta}{\theta^2}\)
From solution 20,
\(\lim_{\theta \to 0} \dfrac{sin \theta}{\theta} \dfrac{sin \theta}{\theta} \dfrac{1}{1 + cos \theta}\)
\(= 1.1.\dfrac{1}{2} = \dfrac{1}{2}\)
22 Solution 22
\(\lim_{\theta \to 0} \dfrac{2 - \sqrt{cos \theta + 3}}{\theta^2}\)
\(= \lim_{\theta \to 0} \dfrac{2 - \sqrt{cos \theta + 3}}{\theta^2} * \dfrac{2 + \sqrt{cos \theta + 3}}{2 + \sqrt{cos \theta + 3}}\)
\(= \lim_{\theta \to 0} \dfrac{4-(cos \theta + 3)}{\theta^2(2 + \sqrt{cos \theta + 3})}\)
\(= \lim_{\theta \to 0} \dfrac{1-cos\theta}{\theta^2(2 + \sqrt{cos \theta + 3})}\)
\(= \lim_{\theta \to 0} \dfrac{1 - cos\theta}{\theta^2} * \lim_{\theta \to 0} \dfrac{1}{2 + \sqrt{cos \theta + 3}}\)
\(= \dfrac{1}{2} * \dfrac{1}{2 + \sqrt{1 + 3}} = \dfrac{1}{2} * \dfrac{1}{4} = \dfrac{1}{8}\)
23 Solution 23
\(\lim_{x \to 0} \dfrac{sin 5x}{5x}\)
Rewriting the above equation as \(\lim{x \to 0} f(g(x))\)
where \(g(x) = 5x\), \(f(x) = \dfrac{sin x}{x}\)
Let \(u = g(x)\) and \(y = f(u)\)
\(\lim_{x \to 0} g(x) = 0\)
As \(x \to 0\), \(u \to 0\)
\(\lim_{u \to 0} f(u) = 1\)
So, \(\lim_{x \to 0} \dfrac{sin 5x}{5x} = 1\)
24 Solution 24
\(\lim_{x \to 0} \dfrac{tan 5x}{3x}\)
\(= \lim_{x \to 0} \dfrac{tan 5x}{3x}\)
\(= \lim_{x \to 0} \dfrac{sin 5x * 5}{cos 5x * 5x * 3}\)
\(= \lim_{x \to 0} \dfrac{sin 5x}{5x} * \lim_{x \to 0} \dfrac{5}{3 cos 5x}\)
\(= 1.\dfrac{5}{3cos 0} = \dfrac{5}{3}\)
25 Solution 25
\(\lim_{x \to 0} \dfrac{tan 5x}{x^2 - 3x}\)
\(= \lim_{x \to 0} \dfrac{tan 5x}{x(x-3)}\)
\(= \lim_{x \to 0} \dfrac{tan 5x}{x(x-3)} * \dfrac{3}{x-3}\)
\(= \lim_{x \to 0} \dfrac{tan 5x}{3x} * \lim_{x \to 0} \dfrac{3}{x-3}\)
\(= \dfrac{5}{3} * \dfrac{3}{-3}\)
\(= \dfrac{-5}{3}\)
26 Solution 26
\(f(x) = \dfrac{x^2 - 2x + 7}{x^3 - 3x^2}\)
\(f(x) = \dfrac{(x-1)^2 + 6}{x^2(x-3)}\)
\(f(x) = \dfrac{(1-\dfrac{1}{x})^2 + \dfrac{6}{x^2}}{x(1-\dfrac{3}{x})}\)
26.1 Solution a
As \(x \to 0\), the numerator tends to \(7\).
As \(x \to 0^-\), the denominator tends to \(0^-\).
As \(x \to 0^+\), the denominator tends to \(0^-\).
We know that as \(x \to 0^-\), \(\dfrac{1}{x} \to -\infty\)
So, \(\lim_{x \to 0} f(x) = -\infty\)
26.2 Solution b
As \(x \to 3^-\), the numerator tends to \(10\).
As \(x \to 3^-\), the denominator tends to \(0^-\).
\(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \dfrac{(1-\dfrac{1}{x})^2 + \dfrac{6}{x^2}}{x(1-\dfrac{3}{x})}\)
\(= -\infty\)
26.3 Solution c
\(\lim_{x \to 3^+} f(x)\)
As \(x \to 3^+\), the numerator tends to \(10\).
As \(x \to 3^+\), the denominator tends to \(0^+\).
\(\lim_{x \to 3^+} f(x) = \infty\)
26.4 Solution d
\(\lim_{x \to \infty} f(x)\)
\(= \lim_{x \to \infty} \dfrac{(1-\dfrac{1}{x})^2 + \dfrac{6}{x^2}}{x(1-\dfrac{3}{x})}\)
\(= 0\)
26.5 Solution e
\(\lim_{x \to -\infty} f(x)\)
\(= \lim_{x \to -\infty} \dfrac{(1-\dfrac{1}{x})^2 + \dfrac{6}{x^2}}{x(1-\dfrac{3}{x})}\)
\(= 0\)
27 Solution 27
27.1 Solution a
We need to prove \(0 \leq a < b \implies a^2 < b^2\)
Suppose \(0 \leq a < b\)
So, \(a < b\)
\(a - b < 0\)
\((a-b)(a+b) < 0\)
\(a^2 - b^2 < 0\)
\(a^2 < b^2\)
So, \(0 \leq a < b \implies a^2 < b^2 \blacksquare\)
27.2 Solution b
Suppose \(r \geq 0\)
Let us assume that there exists two numbers \(c_1\) and \(c_2\) such that
\(c_1 \neq c_2\)
\(c_1 \geq 0\)
\(c_2 \geq 0\)
\(c_1^2 = r\)
\(c_2^2 = r\)
Since \(c_1^2 = r\) and \(c_2^2 = r\).
So, \(c_1^2 = c_2^r\)
Taking square root on both sides, \(c_1 = c_2\) since both \(c_1 \geq 0\) and \(c_2 \geq 0\).
But this contradicts with our initial assumption that \(c_1 \neq c_2\).
So, there is exactly one number \(c \geq 0\) such that \(c^2 = r\).
27.3 Solution c
We need to prove \(0 \leq a < b \implies \sqrt{a} < \sqrt{b}\)
Suppose \(0 \leq a < b\)
So, \(a < b\)
Since \(a \geq 0\) and \(b > 0\)
\(\sqrt{a} < \sqrt{b}\)
So, \(0 \leq a < b \implies \sqrt{a} < \sqrt{b} \blacksquare\)
28 Solution 28
We need to prove that \(\lim_{x \to a^-} \sqrt{x} = \sqrt{a}\)
Suppose \(a > 0\)
We will use the \(\epsilon - \delta\) method.
Suppose \(\epsilon > 0\). Let \(b = (\sqrt{a} - \epsilon)^2\) and \(\delta = a - b\)
Since \(b = (\sqrt{a} - \epsilon)^2\)
\(\sqrt{b} = \sqrt{a} - \epsilon\)
Since \(0 < a\), \(0 < \sqrt{a}\)
So, \(0 < \sqrt{a} - \epsilon < \sqrt{a}\)
\(0 < \sqrt{b} < \sqrt{a}\)
Suppose \(a - \delta < x < a\)
We need to prove that \(|\sqrt{x} - \sqrt{a}| < \epsilon\)
From \(\delta = a - b\), \(b = a - \delta\)
So, \(b < x < a\)
From \(0 < \sqrt{b}\), \(0 < b\). So, \(b > 0\)
Since \(a > 0\) and \(b > 0\)
So, \(\sqrt{b} < \sqrt{x} < \sqrt{a}\)
\(\sqrt{a} - \epsilon < \sqrt{x} < \sqrt{a}\)
and therefore \(|\sqrt{x} - \sqrt{a}| < \epsilon\) as required.
29 Solution 29
Let \(f(x) = sin x\)
where \(x\) is in degrees.
We need to find \(\lim_{x \to 0} \dfrac{f(x)}{x}\)
Let \(g(x)\) be the number of radians in angle of \(x\) degree.
So, \(\lim_{x \to 0} \dfrac{sin(g(x))}{x}\)
\(360\) degree = \(2\pi\) radians
\(1\) degree = \(\dfrac{\pi}{180}\) radians
\(\lim_{x \to 0} \dfrac{sin(\dfrac{\pi*x}{180})}{x}\)
\(= \lim_{x \to 0} \dfrac{sin(\dfrac{\pi*x}{180})}{(\pi * \dfrac{x}{180}) * \dfrac{180}{\pi}}\)
\(= \dfrac{\pi}{180} \lim_{x \to 0} \dfrac{ sin(\dfrac{\pi * x}{180})}{\dfrac{\pi * x}{180}}\)
\(Let y = \dfrac{\pi * x}{180}\)
As \(x \to 0\), \(y \to 0\)
So, \(\lim_{y \to 0} \dfrac{\pi}{180} * \dfrac{sin(y)}{y} = \dfrac{\pi}{180}\)
30 Solution 30
30.1 Solution a
Suppose as \(x \to a^{\neq}\), \(g(x) \to L^{\leq}\)
\(f\) is continous from left at \(a\).
From theorem 2.7.12, as \(x \to a^{\leq}\), \(f(x) \to f(a)\)
We need to prove \(\lim_{x \to a} f(g(x)) = f(L)\)
Let \(u = g(x)\)
Since \(\lim_{x \to a} g(x) = L\), we can say that as \(x \to a^{\neq}\), \(u \to L^{\leq}\)
Applying theorem 2.7.12,
As \(u \to L^{\leq}\), \(f(u) \to f(L)\)
So, \(\lim_{x \to a} f(g(x)) = f(L)\).
30.2 Solution b
Suppose as \(x \to a^{\neq}\), \(g(x) \to L^{\geq}\)
We need to prove \(\lim_{x \to a} f(g(x)) = f(L)\)
Let \(u = g(x)\)
As \(x \to a^{\neq}\), \(g(x) \to L^{\geq}\)
Since \(f\) is continous from the right at \(L\), from theorem 2.7.12,
As \(u \to L^{\geq}\), \(f(u) \to f(L)\)
So, \(\lim_{x \to a} f(g(x)) = f(L)\)
31 Solution 31
\(f\) is a function
\(I\) is an interval contained in the domain of \(f\).
\(a \in I\)
31.1 Solution a
Suppose \(f\) is continous on an interval \(I\).
Suppose \(a \in I\).
Since \(f\) is continous, \(\lim_{x \to a} f(x) = f(a)\)
From above, \(\forall \epsilon > 0 \exists \delta > 0 |x - a | < \delta \implies |f(x) - L| < \epsilon\)
Now let's prove \(x \to a^{\in I}\), \(f(x) \to f(a)\)
Suppose \(\epsilon_1 > 0\). Suppose \(\delta_1 > 0\)
Suppose \(|x-a| < \delta_1\) and \(x \in I\)
From above, we can conclude \(|f(x) - L| < \epsilon_1\)
Hence \(x \in a^{\in I}\), \(f(x) -> f(a)\)
Now we have to prove the other side of the equivalence.
\(\forall \epsilon > 0 \exists \delta > 0 |x-a| < \delta \land x \in I \implies |f(x) - L| < \epsilon\)
Suppose \(\epsilon_1 > 0\). Suppose \(\delta_1 > 0\)
Suppose \(|x-a| < \delta_1\)
Since \(f\) is continous on an interval \(I\), we know that \(x \in I\).
So, \(|f(x)-L| < \epsilon_1\)
31.2 Solution b
\(g\) is continous on an interval \(J\).
\(f\) is continous on an interval \(I\).
We need to prove
\(\forall a \in J x \to a^{\in J}, g(x) \to g(a)^{\in I}\)
\(\forall \epsilon > 0 \exists \delta > 0 |x-a| < \delta \land x \in J \implies |g(x) - g(a)| < \epsilon \land g(x) \in I\)
Suppose \(\epsilon > 0, \delta > 0\) and \(|x-a| < \delta \land x \in J\)
Since \(g\) is continous on \(J\),
\(\lim_{x \to a} g(x) = g(a)\)
So, \(|g(x) - g(a)| < \epsilon\)
The only remaining thing to prove is \(g(x) \in I\). We know that \(x \in J\).
Also from theorem 2.7.13,
\(\forall x \in J, g(x) \in I\).
So, \(g(x) \in I \blacksquare\)
31.3 Solution c
Suppose \(a \in J\). Then by the second part of the qeustion, we know that \(g(a) \in I\). Now we know that \(f\) is continous on an interval \(I\). So, \(f(g(a))\) is continous on interval \(I\). So, \(f \circ g\) is continous on \(J\).
32 Solution 32
32.1 Solution b
Suppose \(-\dfrac{\pi}{2} < \theta < 0\)
So, \(0 < -\theta < \dfrac{\pi}{2}\)
Suppose \(cos (-\theta) \leq \dfrac{sin(-\theta)}{-\theta} \leq 1\)
Since \(cos (-\theta) = cos(\theta)\)
\(sin(-\theta) = -sin(\theta)\)
So, \(cos(\theta) \leq \dfrac{sin(\theta)}{\theta} \leq 1\)