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granice

: 09 maja 2017, 19:32
autor: mochel
oblicz
a)\(\Lim_{x\to- \infty } \frac{-2x+4x^3+6x^7- \frac{5}{x^8} }{-9- \frac{4}{x}+7x^4 }\)
b)\(\Lim_{n\to \infty } ( \sqrt{n^2-2n-1} - \sqrt{3-7n+n^2} )\)
c)\(\Lim_{x\to 2 } \frac{4-x^2}{2x^2-3x+1}\)
d)\(\Lim_{n\to \infty } ( \frac{8n^2+3}{8n^2} )^{5n^2}\)
e)\(\Lim_{x\to 0} \frac{ \tg 2x}{5x}\)

Re: granice

: 09 maja 2017, 19:43
autor: eresh
mochel pisze:oblicz
a)\(\Lim_{x\to- \infty } \frac{-2x+4x^3+6x^7- \frac{5}{x^8} }{-9- \frac{4}{x}+7x^4 }\)

\(\Lim_{x\to- \infty } \frac{-2x+4x^3+6x^7- \frac{5}{x^8} }{-9- \frac{4}{x}+7x^4 }=
\Lim_{x\to -\infty}\frac{\frac{-2}{x^3}+4+6x^3-\frac{5}{x^{12}}}{\frac{-9}{x^4}-\frac{4}{x^5}+7}=\frac{-\infty}{7}=-\infty\)

Re: granice

: 09 maja 2017, 19:46
autor: eresh
mochel pisze:oblicz

b)\(\Lim_{n\to \infty } ( \sqrt{n^2-2n-1} - \sqrt{3-7n+n^2} )\)

\(\Lim_{n\to\infty}\frac{n^2-2n-1-3+7n-n^2}{ ( \sqrt{n^2-2n-1} + \sqrt{3-7n+n^2} )}=\Lim_{n\to\infty}\frac{n(5-\frac{4}{n})}{n(\sqrt{1-\frac{2}{n}-\frac{1}{n^2}}+\sqrt{\frac{3}{n^2}-\frac{7}{n}+1})}=\Lim_{n\to\infty}\frac{5-\frac{4}{n}}{\sqrt{1-\frac{2}{n}-\frac{1}{n^2}}+\sqrt{\frac{3}{n^2}-\frac{7}{n}+1}}=\frac{5}{2}\)

Re: granice

: 09 maja 2017, 19:48
autor: eresh
mochel pisze:oblicz
c)\(\Lim_{x\to 2 } \frac{4-x^2}{2x^2-3x+1}\)
\\(Lim_{x\to 2}\frac{4-x^2}{2x^2-3x+1}=\frac{0}{3}=0\)

Re: granice

: 09 maja 2017, 19:52
autor: eresh
mochel pisze:oblicz

d)\(\Lim_{n\to \infty } ( \frac{8n^2+3}{8n^2} )^{5n^2}\)

\(\Lim_{n\to \infty } \left( \frac{8n^2+3}{8n^2} \right)^{5n^2}=\Lim_{n\to\infty} \left(1+\frac{3}{8n^2} \right)^{5n^2}=\Lim_{n\to\infty} \left[ \left(1+\frac{3}{8n^2} \right)^{8n^2}\right]^{\frac{5n^2}{8n^2}}=(e^{3})^{\frac{5}{8}}=e^{\frac{15}{8}}\)

Re: granice

: 09 maja 2017, 19:57
autor: eresh
mochel pisze:oblicz

e)\(\Lim_{x\to 0} \frac{ \tg 2x}{5x}\)

\(\Lim_{x\to 0}\frac{\tg 2x}{5x}=\Lim_{x\to 0}(\frac{\sin 2x}{\cos 2x}\cdot\frac{1}{5x})=\Lim_{x\to 0}(\frac{\sin 2x}{2x}\cdot\frac{\cos 2x}{\frac{5}{2}})=1\cdot\frac{2}{5}=\frac{2}{5}\)

Re: granice

: 14 maja 2017, 14:28
autor: mochel
eresh pisze:
mochel pisze:oblicz

e)\(\Lim_{x\to 0} \frac{ \tg 2x}{5x}\)

\(\Lim_{x\to 0}\frac{\tg 2x}{5x}=..=\Lim_{x\to 0}(\frac{\sin 2x}{2x}\cdot\frac{\cos 2x}{\frac{5}{2}})=..\)
czy mogę prosić o większe rozpisanie skąd się to wzięło?

: 14 maja 2017, 16:31
autor: Galen
\(\frac{tg2x}{5x}=\frac{ \frac{sin2x}{cos2x} }{2,5 \cdot 2x}= \frac{sin2x}{2x} \cdot \frac{\frac{1}{cos2x}}{2,5} \to 1 \cdot \frac{1}{2,5}= \frac{1}{ \frac{5}{2} }= \frac{2}{5}\)