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% !!! IMAGES START HERE !!!

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$\displaystyle \lim_{x \rightarrow 0} \frac{ arcsen(5x)}{3x^2-4x}$%
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{\newpage\clearpage
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$\displaystyle \lim_{x \rightarrow +\infty} \left( 10x^2+3x-4\right)^{\frac{1}{
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{\newpage\clearpage
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$0/0$%
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{\newpage\clearpage
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$\displaystyle \lim_{x\rightarrow 0} \frac{ arcsen(5x)}{3x^2-4x} =
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$\infty/\infty$%
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$\displaystyle L = \lim_{x \rightarrow +\infty} 
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$\displaystyle \ln(L) = \ln \left( \lim_{x \rightarrow +\infty}
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$ln(x)$%
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{\newpage\clearpage
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$\displaystyle \ln(L) = \lim_{x \rightarrow +\infty}
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$\infty/\infty \Rightarrow$%
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{\newpage\clearpage
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$\displaystyle \ln(L) = \lim_{x\rightarrow +\infty}
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$\displaystyle \ln(L) = \lim_{x\rightarrow +\infty}
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{\newpage\clearpage
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$ \displaystyle \lim_{x \rightarrow +\infty}
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$f$%
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$(0,0)$%
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$(-\infty,0)$%
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$(0,+\infty)$%
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$\displaystyle f(x) = \frac{5x^2}{x^2+3}$%
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$f''(x) > 0$%
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$(-1,1)$%
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$f''(x) < 0$%
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$(-\infty,-1) \bigcup (1,+\infty)$%
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{\newpage\clearpage
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$\displaystyle f'(x) = \frac{(x^2+3)(10x) - 5x^2 (2x)}{(x^2+3)^2} =
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$f(x)$%
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$(-\infty,0]$%
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$[0,+\infty)$%
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$-\infty$%
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$\displaystyle \lim_{x \rightarrow -\infty} 
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$y=5$%
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$+\infty$%
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{\newpage\clearpage
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$\displaystyle \lim_{x \rightarrow +\infty} 
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$(-\infty,+\infty)$%
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{\newpage\clearpage
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$f'(x)$%
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{\newpage\clearpage
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$D(f') = (-\infty,+\infty)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline444}%
$\displaystyle f(0) = \frac{5(0)}{0+3} = 0 \ , \
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\lthtmlinlinemathZ
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{\newpage\clearpage
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$x$%
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{\newpage\clearpage
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$y = 16-x^2$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline452}%
$\displaystyle A(x) = \frac{(2x) y}{2} = \frac{2x (16-x^2)}{2}=
x (16-x^2)$%
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{\newpage\clearpage
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$0 < x < 4$%
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{\newpage\clearpage
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$\displaystyle A'(x) = (1)(16-x^2) + x(-2x) = 16 - 3x^2$%
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{\newpage\clearpage
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$16-3x^2 = 0 \Rightarrow x^2 = 16/3 \Rightarrow
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$x = 4/\sqrt{3}$%
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{\newpage\clearpage
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$(0,4)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline468}%
$\displaystyle
A \left(\frac{4}{\sqrt{3}} \right) = \frac{4}{\sqrt{3}}  \cdot
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\frac{128\sqrt{3}}{9}$%
\lthtmlinlinemathZ
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline470}%
$\displaystyle f(x) = x - \frac{\ln(x)}{5}$%
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{\newpage\clearpage
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$\displaystyle x > \frac{\ln(x)}{5}$%
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{\newpage\clearpage
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$x > 0$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline480}%
$\displaystyle \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}}
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{\newpage\clearpage
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$\ln(x) \rightarrow -\infty$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline486}%
$\displaystyle f'(x) = 1 - \frac{1}{5x} \ ,  x > 0$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline488}%
$\displaystyle f''(x) = \frac{1}{5x^2} \ , \ x > 0$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline492}%
$f'(x) = 0 \Leftrightarrow
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline494}%
$f''(x) > 0 \ \ \forall x \in (0,+\infty)$%
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{\newpage\clearpage
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$f''(1/5) > 0$%
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{\newpage\clearpage
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$x=1/5$%
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{\newpage\clearpage
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$f(x) \geq f(1/5)$%
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline512}%
$\displaystyle f \left( \frac{1}{5} \right) = \frac{1}{5} - \frac{\ln(1/5)}{5}
= \frac{1}{5} + \frac{\ln(5)}{5} > 0$%
\lthtmlinlinemathZ
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{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline514}%
$\displaystyle f(x) \geq f(1/5) > 0 \Rightarrow
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{\newpage\clearpage
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$\displaystyle \int x e^{kx}dx$%
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{\newpage\clearpage
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$k$%
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{\newpage\clearpage
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$\displaystyle \int \frac{t}{3t^2+6} dt$%
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{\newpage\clearpage
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$k=0$%
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{\newpage\clearpage
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$\displaystyle \int x e^{kx} dx = \int x (1) dx = \frac{x^2}{2} + C, C \in {\mathbb{R}}$%
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{\newpage\clearpage
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$k \neq 0$%
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{\newpage\clearpage
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$u = x$%
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{\newpage\clearpage
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$\Rightarrow$%
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$ du = dx $%
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\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline536}%
$dv = e^{kx} dx$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline540}%
$\displaystyle v = \frac{e^{kx}}{k}$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline542}%
$\displaystyle \int x e^{kx}dx = \frac{x e^{kx}}{k} - \int \frac{e^{kx}dx}{k}
= \frac{x e^{kx}}{k} - \frac{e^{kx}}{k^2} + C, C \in {\mathbb{R}}$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline544}%
$u = 3t^2 + 6 \Rightarrow du/dt = 6t \Rightarrow du = 6 t \ dt$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}

{\newpage\clearpage
\lthtmlinlinemathA{tex2html_wrap_inline546}%
$\displaystyle \int \frac{t dt}{3t^2 + 6 } = \int \frac{du/6}{u} =
\frac{1}{6} \int \frac{du}{u} = \frac{\ln |u|}{6} + C = 
\frac{\ln | 3t^2+ 6|}{6} + C, C \in {\mathbb{R}}$%
\lthtmlinlinemathZ
\lthtmlcheckvsize\clearpage}


\end{document}