تمرینات-محاسبه حد توابع مثلثاتی و رفع ابهام
حد توابع مثلثاتی را محاسبه کنید و در صورت نیاز رفع ابهام کنید .
[math]1)\mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{\sin 3x}} = ? \\ \mathop {\lim }\limits_{x \to 0} \frac{{4x}}{{\sin 3x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3 \times 4x}}{{3 \times \sin 3x}} = \frac{4}{3}\mathop {\lim }\limits_{x \to 0} \frac{{3x}}{{\sin 3x}} = \frac{4}{3}\mathop {\lim }\limits_{x \to 0} \frac{1}{{\frac{{\sin 3x}}{{3x}}}} = \frac{4}{3}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\mathop {\lim }\limits_{x \to 0}\frac{{\sin 3x}}{{3x}}}\limits_{} }} \\ \left\{ {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}} = 1} \right\} \\ \\ \frac{4}{3}\frac{{\mathop {\lim }\limits_{x \to 0} 1}}{{\mathop {\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x}}{{3x}}}\limits_{} }} = \frac{4}{3} \\[/math]
در مثال بالا از هم ارزیها برای محاسبه حد استفاده کردیم.
[math]2)\mathop {\lim }\limits_{x \to 0} \frac{{\cos 3x – \cos x}}{{{x^2}}} = ? \\[/math]
جواب این حد صفر صفرم و مبهم خواهد یود که باید رفع ابهام کنیم ،در اینجا ما با استفاده از اتحادها رفع ابهام می کنیم.
[math]\cos 3x – \cos x = – 2\sin \frac{{3x – x}}{2}\sin \frac{{3x + x}}{2} = – 2\sin x\sin 2x \\ \mathop {\lim }\limits_{x \to 0} \frac{{\cos 3x – \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{ – 2\sin x\sin 2x}}{{{x^2}}} = – 2\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{x} = – 2 \times 1 \times \mathop {\lim }\limits_{x \to 0} \frac{{2\sin 2x}}{{2x}} \\ – 2 \times 2 \times \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}} = – 4 \\[/math]
[math]3)\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = ? \\ \\ \mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{\sin bx}}.\frac{a}{b}.\frac{{bx}}{{ax}}} \right) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin ax}}{{ax}}.\frac{{bx}}{{\sin bx}}.\frac{a}{b}} \right) = \frac{a}{b}\frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}}}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}} = \frac{a}{b}.\frac{1}{1} = \frac{a}{b} \\[/math]
[math]4)\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – \sin x}}{{{x^3}}} = ? \\ \\ \mathop {\lim }\limits_{x \to 0} \frac{{\tan x – \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin x}}{{\cos x}} – \sin x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x(\frac{1}{{\cos x}} – 1)}}{{{x^3}\cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x(1 – \cos x)}}{{{x^3}}} = \\ \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin x}}{x}.\frac{{1 – \cos x}}{{{x^2}\cos x}}} \right] \\[/math]
از اتحادهای مثلثاتی می دانیم که :
[math] 1 – \cos x = 2{\sin ^2}\frac{x}{2}[/math]
[math]\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin x}}{x}.\frac{{1 – \cos x}}{{{x^2}\cos x}}} \right] = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin x}}{x}.\frac{{2{{\sin }^2}\frac{x}{2}}}{{{x^2}\cos x}}} \right] = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.2\mathop {\lim }\limits_{x \to 0} {{\sin }^2}\frac{x}{2}}}{{\mathop {\lim }\limits_{x \to 0} \cos x}}[/math]
می دانیم که تساویهای زیر برقرار است :
[math]\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1,\mathop {\lim }\limits_{x \to 0} \cos x = 1 \\\\2\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}} = 2\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sin }^2}\frac{x}{2}}}{{{x^2}}}.\frac{4}{4}} \right) = 2\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{{\sin }^2}\frac{x}{2}}}{{\frac{{{x^2}}}{4}}}.\frac{1}{4}} \right) = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}\frac{x}{2}}}{{{{(\frac{x}{2})}^2}}} = \frac{1}{2} \\[/math]
[math]5)\mathop {\lim }\limits_{x \to a} \frac{{\sin (x – a)}}{{{x^2} – {a^2}}} = ? \\[/math]
چون حد مبهم است پس سعی می کنیم با استفاده از تغییر متغیر آن را حل کنیم :
[math]x – a = t \Rightarrow x = t + a \\ x \to a \Rightarrow t \to 0 \\ \mathop {\lim }\limits_{x \to a} \frac{{\sin (x – a)}}{{{x^2} – {a^2}}} = \mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{{(t + a + a)t}} = \mathop {\lim }\limits_{t \to 0} \frac{1}{{t + 2a}}\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} = \frac{1}{{2a}} \times 1 = \frac{1}{{2a}} \\[/math]
[math]6)\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\cos (x + \frac{\pi }{4})}}{{\tan x – 1}} = ? \\ [/math]
برای حل این مساله باید از تغییر متغیر استفاده کنیم
[math] x – \frac{\pi }{4} = t \Rightarrow x \to \frac{\pi }{4} \Rightarrow t \to 0 \\ \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\cos (x + \frac{\pi }{4})}}{{\tan x – 1}} = \mathop {\lim }\limits_{t \to 0} \frac{{\cos (\frac{\pi }{4} + t + \frac{\pi }{4})}}{{\tan (\frac{\pi }{4} + t) – 1}} = \mathop {\lim }\limits_{t \to 0} \frac{{\cos (\frac{\pi }{2} + t)}}{{\tan (\frac{\pi }{4} + t) – 1}} = \\\mathop {\lim }\limits_{t \to 0} \frac{{ – \sin t}}{{\frac{{1 + \tan t}}{{1 – \tan t}} – 1}} = \\[/math]
اکنون برای سینوس و تانژانت از هم ارزیهای آنها استفاده می کنیم و خواهیم داشت که :
[math]\mathop {\lim }\limits_{t \to 0} \frac{{ – \sin t}}{{\frac{{1 + \tan t}}{{1 – \tan t}} – 1}} = \mathop {\lim }\limits_{t \to 0} \frac{{ – t}}{{\frac{{1 + t}}{{1 – t}} – 1}} = \mathop {\lim }\limits_{t \to 0} \frac{{ – t}}{{\frac{{1 + t – 1 + t}}{{1 – t}}}} = \mathop {\lim }\limits_{t \to 0} \frac{{ – t(1 – t)}}{{2t}} = \mathop {\lim }\limits_{t \to 0} \frac{{t – 1}}{2} = \frac{{ – 1}}{2}[/math]
[math]7)\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} – 2}}{{\sin (x – 1)}} = ? \\ \\ \left\{ \begin{array}{l} {x^3} + {x^2} – 2 = (x – 1)({x^2} + 2x + 2) \\ \mathop {\lim }\limits_{x \to 1} \frac{{{x^3} + {x^2} – 2}}{{\sin (x – 1)}} \\ \end{array} \right\} \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{(x – 1)({x^2} + 2x + 2)}}{{\sin (x – 1)}} = \\ \mathop {\lim }\limits_{x \to 1} \frac{{x – 1}}{{\sin (x – 1)}} \times \mathop {\lim }\limits_{x \to 1} ({x^2} + 2x + 2) = \left\{ {\mathop {\lim }\limits_{x \to 1} \frac{{x – 1}}{{\sin (x – 1)}} = 1} \right\} \\ = 1 \times \mathop {\lim }\limits_{x \to 1} ({x^2} + 2x + 2) = 5 \\[/math]
[math]8)\mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\sin x – 1){\tan ^2}x = ? \\ \left\{ \begin{array}{l} x \to \frac{\pi }{2} \Rightarrow x – \frac{\pi }{2} = t \Rightarrow t \to 0 \\ \mathop {\lim }\limits_{x \to \frac{\pi }{2}} (\sin x – 1){\tan ^2}x = \\ \end{array} \right\} \Rightarrow \mathop {\lim }\limits_{t \to 0} (\sin (\frac{\pi }{2} + t) – 1)({\tan ^2}(\frac{\pi }{2} + t)) = \\ \\ \mathop {\lim }\limits_{t \to 0} (\cos t – 1)({\cot ^2}t) = \mathop {\lim }\limits_{t \to 0} \frac{{ – (1 – \cos t)}}{{{{\tan }^2}t}} = \mathop {\lim }\limits_{t \to 0} \frac{{ – (\frac{1}{2}{t^2})}}{{{t^2}}} = \frac{{ – 1}}{2} \\ [/math]