Найти предел, стремящийся к бесконечности (5x+3)*[ln(x-2)-ln(x+1)]

1.

\begin{gathered} < var > \\y=e-\ln x\\ y'=-\frac{1}{x}\\ < /var > \end{gathered}

<var>

y=e−lnx

y

′

=−

x

1

</var>

2.

\begin{gathered} < var > \\y=\ln(10-5x)\\ y'=\frac{1}{10-5x}\cdot-5\\ y'=\frac{-5}{10-5x}\\ y'=\frac{-5}{5(2-x)}\\ y'=\frac{1}{x-2} < /var > \end{gathered}

<var>

y=ln(10−5x)

y

′

=

10−5x

1

⋅−5

y

′

=

10−5x

−5

y

′

=

5(2−x)

−5

y

′

=

x−2

1

</var>

3.

\begin{gathered} < var > \\y=3-4\ln (1-x)\\ y'=-4\cdot\frac{1}{1-x}\cdot(-1)\\ y'=-\frac{4}{x-1} < /var > \end{gathered}

<var>

y=3−4ln(1−x)

y

′

=−4⋅

1−x

1

⋅(−1)

y

′

=−

x−1

4

</var>

4.

\begin{gathered} < var > \\y=\ln \frac{1}{x}\\ y'=\frac{1}{\frac{1}{x}}\cdot(-\frac{1}{x^2})\\ y'=-\frac{x}{x^2}\\ y'=-\frac{1}{x} < /var > \end{gathered}

<var>

y=ln

x

1

y

′

=

x

1

1

⋅(−

x

2

1

)

y

′

=−

x

2

x

y

′

=−

x

1

</var>

5.

\begin{gathered} < var > \\y=1-3^x\\ y'=-3^x \ln 3 < /var > \end{gathered}

<var>

y=1−3

x

y

′

=−3

x

ln3</var>