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[kaum_erdbeere]

czesc, mógłby mi ktos z tym pomoc?:(

\(a) \lim_{x\to 0} \frac{cosx-cos5x}{x^2}

b) \lim_{x\to4 } \frac{ \sqrt{1+2x}-3 }{ \sqrt{x} -2}

c) \lim_{x\to- \infty } (\frac{x^2-1}{x^2+1})^(2x^2)

d) \lim_{x\to \infty } x( \sqrt{x^2+1} -x)

e) \lim_{x\to0 } \frac{1-cosx}{x}

f) \lim_{x\to 0 } \frac{ \sqrt{^2+1}- \sqrt{x+1} }{1- \sqrt{x+1} }\)

[Galen]

a)
Zastosuj dwukrotnie De L'Hosopitala
\(\lim_{x\to 0} \frac{cosx-cos5x}{x^2}=(H)= \lim_{x\to 0} \frac{-sinx+5sin5x}{2x}=(H)= \lim_{x\to 0} \frac{cosx=25sos5x}{2}=\frac{24}{2}=12\)

[kaum_erdbeere]

problem w tym, ze nie bylo tego twierdzenia

[radagast]

No to tak : \(a) \lim_{x\to 0} \frac{cosx-cos5x}{x^2}=^* \lim_{x\to 0} \frac{2sin 3x sin2x}{x^2}=2\lim_{x\to 0} \frac{sin 3x }{x} \cdot \frac{sin2x}{x}=2\lim_{x\to 0} 3\frac{sin 3x }{3x} \cdot 2\frac{sin2x}{2x}=12\)
\(*\) ze wzoru na różnicę cosinusów

[Galen]

b)
\(\lim_{x\to 4} \frac{(\sqrt{1+2x}-3)(\sqrt{1+2x}+3)(\sqrt{x}+2))}{(\sqrt{x}-2)(\sqrt{x}+2)(\sqrt{1+2x}+3)}= \lim_{x\to 4} \frac{(1+2x-9)(\sqrt{x}+2)}{(x-4)(\sqrt{1+2x}+3)}= \lim_{x\to 4} \frac{2(x-4)(\sqrt{x}+2)}{(x-4)(\sqrt{1+2x}+3)} =\frac{8}{6}= \frac{4}{3}\)

[Galen]

d)
\(= \lim_{x\to \infty } \frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}= \lim_{x\to \infty } \frac{x}{x(\sqrt{1+\frac{1}{x^2}+1})}=\frac{1}{2}\)
e)
\(\lim_{x\to 0} \frac{1-cos x}{x}\cdot \frac{1+cosx}{1+cosx}= \lim_{x\to 0} \frac{1-cos^2x}{x(1+cosx)}= \lim_{x\to 0} \frac{sin^2x}{x(1+cosx)}=\)

\(= \lim_{x\to 0} \frac{sinx}{x} \cdot \lim_{x\to 0} \frac{sinx}{1+cosx}=1 \cdot \frac{0}{2}=0\)

[Galen]

c)
\(\lim_{x\to - \infty } (\frac{x^2-1}{x^2+1})^{2x^2}= \lim_{x\to - \infty }( \frac{x^2(1- \frac{1}{x^2} )}{x^2(1+ \frac{1}{x^2} )} )^{2x^2}= \lim_{x\to - \infty } \frac{((1+ \frac{-1}{x^2} )^{x^2})^2}{((1+ \frac{1}{x^2} )^{x^2})^2}=\)

\(= \frac{(e^{-1})^2}{e^2}= \frac{e^{-2}}{e^2} =e^{-4}= \frac{1}{e^4}\)