limes

[mol_ksiazkowy]

\(\displaystyle{ \lim_{x \to a^+} [\frac{ln(sin(2x-2a))}{ln(sin(x-a))}]^{\frac{1}{x-a}}=?}\)

[max]

Niech \(\displaystyle{ t = x - a}\)
Wtedy:
\(\displaystyle{ \frac{\ln \sin 2t}{\ln \sin t} = \frac{\ln (2\sin t\cos t)}{\ln \sin t} = 1 + \frac{\ln (2\cos t)}{\ln \sin t}\\
\lim_{t\to 0^{+}}\left(1 + \frac{\ln (2\cos t)}{\ln \sin t}\right)^{\frac{1}{t}} =\\
= \lim_{t\to 0^{+}} ft(\left(1 + \frac{\ln (2\cos t)}{\ln \sin t}\right)^{\frac{\ln \sin t}{\ln (2\cos t)}}\right)^{\frac{\ln (2\cos t)}{\sin t \ln \sin t}\cdot \frac{\sin t}{t}} =\\
= \lim_{u\to -\infty}e^{u} = 0}\)

Korzystamy z tego, że:
\(\displaystyle{ \lim_{\alpha \to 0} (1 + )^{1/\alpha} = e\\
\lim_{\alpha\to 0} \frac{\sin }{\alpha} = 1\\
\lim_{\alpha\to 0^{+}}\alpha \ln = 0^{-}}\)
oraz z ciągłości funkcji elementarnych.

[mol_ksiazkowy]

\(\displaystyle{ lim_{t \to 0} \frac{\ln (2\cos t)}{\sin t \ln \sin t} =?}\)

[max]

\(\displaystyle{ \lim_{t\to 0^{+}}\frac{\ln (2\cos t)}{\sin t \ln \sin t}=\left[\frac{\ln 2}{0^{-}}\right] = -\infty}\)