Odpowiedź:
z.2
a ) [tex]\lim_{x \to 3^+} \frac{2 x + 1}{9 - x^{2} } =[/tex] - ∞
b) [tex]a_n = \frac{n^2 +1 - ( n^2 + 5)}{\sqrt{n^2 + 1} +\sqrt{n^2 + 5} } = \frac{- 4}{\sqrt{n^2 +1} + \sqrt{n^2 + 5} }[/tex]
więc [tex]\lim_{n \to \infty} a_n = 0[/tex]
c )
f ( x ) = [ ( 1 + [tex]\frac{3}{8 x} )^{8x}]^{3/4}[/tex]
więc
[tex]\lim_{x \to \infty} f( x) = [ e^3 ]^{3/4} = e^{9/4} = e^{2,25}[/tex]
z.3 a)
f ' (x ) = 4 x³*tg x + [tex]x^4*\frac{1}{cos^2 x} = 4 x^3*tg x + \frac{x^4}{cos^2 x}[/tex]
f ' ( x ) = [tex]\frac{\frac{1}{x} * sin x - ( ln x)* cos x}{sin^2 x} = \frac{\frac{sin x}{x} - ( ln x)* cos x}{sin^2 x}[/tex]
b)
f ( x ) = [tex]e^{10 x}* x + \pi[/tex]
f '( x) = 10*[tex]e^{10 x}[/tex] * x + [tex]e^{10 x}[/tex] = ( 10 x + 1 )*[tex]e^{10 x}[/tex]
f ' ( x) = 0 ⇔ x = - 0,1 bo [tex]e^{10x} > 0[/tex] dla x ∈ R
f '' ( x) = 10*[tex]e^{10 x} +[/tex] ( 10 x + 1)*10*[tex]e^{10 x}[/tex]
Mamy f '' ( - 0, 1) > 0
więc funkcja f ma w x = - 0,1 minimum lokalne.
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Odpowiedź:
z.2
a ) [tex]\lim_{x \to 3^+} \frac{2 x + 1}{9 - x^{2} } =[/tex] - ∞
b) [tex]a_n = \frac{n^2 +1 - ( n^2 + 5)}{\sqrt{n^2 + 1} +\sqrt{n^2 + 5} } = \frac{- 4}{\sqrt{n^2 +1} + \sqrt{n^2 + 5} }[/tex]
więc [tex]\lim_{n \to \infty} a_n = 0[/tex]
c )
f ( x ) = [ ( 1 + [tex]\frac{3}{8 x} )^{8x}]^{3/4}[/tex]
więc
[tex]\lim_{x \to \infty} f( x) = [ e^3 ]^{3/4} = e^{9/4} = e^{2,25}[/tex]
z.3 a)
f ' (x ) = 4 x³*tg x + [tex]x^4*\frac{1}{cos^2 x} = 4 x^3*tg x + \frac{x^4}{cos^2 x}[/tex]
f ' ( x ) = [tex]\frac{\frac{1}{x} * sin x - ( ln x)* cos x}{sin^2 x} = \frac{\frac{sin x}{x} - ( ln x)* cos x}{sin^2 x}[/tex]
b)
f ( x ) = [tex]e^{10 x}* x + \pi[/tex]
f '( x) = 10*[tex]e^{10 x}[/tex] * x + [tex]e^{10 x}[/tex] = ( 10 x + 1 )*[tex]e^{10 x}[/tex]
f ' ( x) = 0 ⇔ x = - 0,1 bo [tex]e^{10x} > 0[/tex] dla x ∈ R
f '' ( x) = 10*[tex]e^{10 x} +[/tex] ( 10 x + 1)*10*[tex]e^{10 x}[/tex]
Mamy f '' ( - 0, 1) > 0
więc funkcja f ma w x = - 0,1 minimum lokalne.
Szczegółowe wyjaśnienie: