- #1

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I have to solve this:

[tex]

\lim_{\substack{s\rightarrow 0^+}} s^4 (\frac{1}{2} ln (s) - \frac{1}{8})

[/tex]

Here is what I did so far:

[tex]

\lim_{\substack{s\rightarrow 0^+}} \frac{s^4}{\frac{1}{\frac{1}{2} ln (s) - \frac{1}{8}} = [/tex]

[tex]

= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{\frac{-\frac{1}{2s}}{(\frac{1}{2} ln (s) - \frac{1}{8})^2}} = [/tex]

[tex]

= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} = [/tex]

[tex]

= \lim_{\substack{s\rightarrow 0^+}} \frac{12s^2}{\frac{2 (\frac{1}{2} ln (s) - \frac{1}{8})^2 + 2 (\frac{1}{2} ln (s) - \frac{1}{8})} {[2s (\frac{1}{2} ln (s) - \frac{1}{8})^2]^2}} = 0 [/tex]

Is this OK? If not, can someone help me please?

[tex]

\lim_{\substack{s\rightarrow 0^+}} s^4 (\frac{1}{2} ln (s) - \frac{1}{8})

[/tex]

Here is what I did so far:

[tex]

\lim_{\substack{s\rightarrow 0^+}} \frac{s^4}{\frac{1}{\frac{1}{2} ln (s) - \frac{1}{8}} = [/tex]

[tex]

= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{\frac{-\frac{1}{2s}}{(\frac{1}{2} ln (s) - \frac{1}{8})^2}} = [/tex]

[tex]

= \lim_{\substack{s\rightarrow 0^+}} \frac{4s^3}{-\frac{1}{2s (\frac{1}{2} ln (s) - \frac{1}{8})^2}} = [/tex]

[tex]

= \lim_{\substack{s\rightarrow 0^+}} \frac{12s^2}{\frac{2 (\frac{1}{2} ln (s) - \frac{1}{8})^2 + 2 (\frac{1}{2} ln (s) - \frac{1}{8})} {[2s (\frac{1}{2} ln (s) - \frac{1}{8})^2]^2}} = 0 [/tex]

Is this OK? If not, can someone help me please?

Last edited: