مثالهای حل شده حد گیری با قاعده هوپیتال
یادگیری نحوه حد گیری با قاعده هوپیتال
مثالهای قاعده هوپیتال
در حد های زیر با استفاده از قاعده هوپیتال حد توابع را حساب کنید.
[math]1)\mathop {\lim }\limits_{x \to 1} \frac{{x – \sqrt x }}{{\sqrt x – 1}} = ? \\ \mathop {\lim }\limits_{x \to 1} \frac{{x – \sqrt x }}{{\sqrt x – 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{1 – \frac{1}{{2\sqrt x }}}}{{\frac{1}{{2\sqrt x }}}} = \frac{{1 – \frac{1}{2}}}{{\frac{1}{2}}} = \frac{{\frac{1}{2}}}{{\frac{1}{2}}} = 1 \\[/math]
2-اگر [math] \mathop {\lim }\limits_{x \to – 3} \frac{{ax + 3a}}{{1 – \sqrt {5x + 16} }} = 2[/math] باشد آنگاه a کدام است ؟ (تست کنکور)
[math]\mathop {\lim }\limits_{x \to – 3} \frac{{ax + 3a}}{{1 – \sqrt {5x + 16} }} = hopital \to \mathop {\lim }\limits_{x \to – 3} \frac{a}{{\frac{{ – 5}}{{2\sqrt {5x + 16} }}}} = \frac{a}{{\frac{{ – 5}}{2}}} = \frac{{2a}}{{ – 5}} \\ \\\frac{{2a}}{{ – 5}} = 2 \to a = – 5 \\[/math]
[math]3)\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax – \sin bx}}{{ax – bx}} = ? \\ \\ \mathop {\lim }\limits_{x \to 0} \frac{{\sin ax – \sin bx}}{{ax – bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{a\cos ax – b\cos bx}}{{a – b}} = \frac{{a\cos 0 – b\cos 0}}{{a – b}} = \frac{{a – b}}{{a – b}} = 1 \\[/math]
[math]4) \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{2^x}}} = ? \\ \\ \mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}}}{{{2^x}}} = \frac{\infty }{\infty } \to hopital \to \mathop {\lim }\limits_{x \to \infty } \frac{{({x^2})’}}{{({2^x})’}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2x}}{{{2^x}\ln x}} = \frac{2}{{\ln 2}}\mathop {\lim }\limits_{x \to \infty } \frac{x}{{{2^x}}} = \frac{\infty }{\infty } \\ \frac{2}{{\ln 2}}\mathop {\lim }\limits_{x \to \infty } \frac{{(x)’}}{{({2^x})’}} = \frac{2}{{\ln 2}}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{2^x}\ln 2}} = \frac{2}{{{{(\ln 2)}^2}}}\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{2^x}}} = \frac{2}{{{{(\ln 2)}^2}}}.0 = 0 \\ [/math]
5-اگر [math] \mathop {\lim }\limits_{x \to 1} \frac{{{x^n} – (n + 1)x + n}}{{{{(x – 1)}^2}}} = 10[/math] آنگاه مقدار [math]n[/math] را بدست آورید.
[math]\mathop {\lim }\limits_{x \to 1} \frac{{{x^n} – (n + 1)x + n}}{{{{(x – 1)}^2}}} = \frac{0}{0} \to hopital \\ \mathop {\lim }\limits_{x \to 1} \frac{{({x^n} – (n + 1)x + n)’}}{{({{(x – 1)}^2})’}} = \mathop {\lim }\limits_{x \to 1} \frac{{n{x^{n – 1}} – (n + 1)}}{{2(x – 1)}} = \frac{0}{0} \\ \mathop {\lim }\limits_{x \to 1} \frac{{(n{x^{n – 1}} – (n + 1))’}}{{(2(x – 1))’}} = \mathop {\lim }\limits_{x \to 1} \frac{{n(n – 1){x^{n – 2}}}}{2} = \frac{{n(n – 1)}}{2} \\ \\ \frac{{n(n – 1)}}{2} = 10 \to n(n – 1) = 20 \to n = 5 \\[/math]
[math]6)\mathop {\lim }\limits_{x \to \frac{1}{4}} \frac{{2 + 2\cos 4\pi x}}{{{{(4x – 1)}^2}}} = ? \\ \mathop {\lim }\limits_{x \to \frac{1}{4}} \frac{{(2 + 2\cos 4\pi x)’}}{{({{(4x – 1)}^2})’}} = \mathop {\lim }\limits_{x \to \frac{1}{4}} \frac{{ – 8\pi \sin 4\pi x}}{{8(4x – 1)}} = \frac{0}{0} \to \mathop {\lim }\limits_{x \to \frac{1}{4}} \frac{{( – 8\pi \sin 4\pi x)’}}{{(8(4x – 1))’}} = \\\mathop {\lim }\limits_{x \to \frac{1}{4}} \frac{{ – 32{\pi ^2}\cos 4\pi x}}{{32}} = {\pi ^2} \\[/math]
[math]7)\mathop {\lim }\limits_{x \to 4} (2 – \sqrt x )\tan \frac{{\pi x}}{8} = ? \\ \\ \mathop {\lim }\limits_{x \to 4} (2 – \sqrt x )\tan \frac{{\pi x}}{8} = \mathop {\lim }\limits_{x \to 4} (2 – \sqrt x )\frac{{\sin \frac{{\pi x}}{8}}}{{\cos \frac{{\pi x}}{8}}} = \mathop {\lim }\limits_{x \to 4} \frac{{(2 – \sqrt x )}}{{\cos \frac{{\pi x}}{8}}} = \frac{0}{0} \to hopital \\ \mathop {\lim }\limits_{x \to 4} (\frac{{(2 – \sqrt x ))’}}{{(\cos \frac{{\pi x}}{8})’}} = \mathop {\lim }\limits_{x \to 4} \frac{{\frac{{ – 1}}{{2\sqrt x }}}}{{\frac{{ – \pi }}{8}(\sin \frac{{\pi x}}{8})}} = \frac{2}{\pi } \\[/math]
[math]8)\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + mx)}^n} – {{(1 + nx)}^m}}}{{{x^2}}} = ? \\ \\ \mathop {\lim }\limits_{x \to 0} \frac{{({{(1 + mx)}^n} – {{(1 + nx)}^m})’}}{{({x^2})’}} = \mathop {\lim }\limits_{x \to 0} \frac{{nm{{(1 + mx)}^{n – 1}} – mn{{(1 + nx)}^{m – 1}}}}{{2x}} = \frac{0}{0} \\\mathop {\lim }\limits_{x \to 0} \frac{{(nm{{(1 + mx)}^{n – 1}} – mn{{(1 + nx)}^{m – 1}})’}}{{(2x)’}} = \mathop {\lim }\limits_{x \to 0} \frac{{{m^2}n(n – 1){{(1 + mx)}^{n – 2}} – m{n^2}(m – 1){{(1 + nx)}^{m – 2}}}}{2} = \\ \frac{{{m^2}n(n – 1) – m{n^2}(m – 1)}}{2} = \frac{{mn(n – m)}}{2} \\[/math]