Limit Problem - # 3

[Jason76]

\(\displaystyle \lim x \rightarrow 0^{+}[\sin x \ln 2x]\)

\(\displaystyle \lim x \rightarrow 0^{+}[\sin (0) \ln 2(0)]\)

\(\displaystyle \lim x \rightarrow 0^{+}[(0)( \ln 0)]\)

\(\displaystyle \lim x \rightarrow 0^{+}[(0)( -\infty)]\) Indeterminate

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\dfrac{1}{\sin x}}] \) Rewrite

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\csc x}] \)

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{d}{dx} \ln x}{\dfrac{d}{dx} \csc x}]\)

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{x}}{-\csc x \cot x}]\)

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{0}}{-\csc 0 \cot 0}]\)

 

[DrPhil]

\(\displaystyle \lim x \rightarrow 0^{+}[\sin x \ln 2x]\)

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\dfrac{1}{\sin x}}] \) Rewrite

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\ln2 x}{\csc x}] \)

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{d}{dx} \ln x}{\dfrac{d}{dx} \csc x}]\)... Missing the "2" in ln(2x)

\(\displaystyle \lim x \rightarrow 0^{+}[\dfrac{\dfrac{1}{x}}{-\csc x \cot x}]\)... but 2/(2x) = 1/x is OK

\(\displaystyle \displaystyle \lim_{x \to 0+} \left[ \dfrac{1/x}{-\csc x \cot x}\right] = \lim_{x \to 0+} \left[ \dfrac{-\sin x}{x \cot x}\right] = - \lim_{x \to 0+}\dfrac{\sin x}{x} = \ \cdot \ \cdot\ \cdot\)

 

[Jason76]

\(\displaystyle \lim x \rightarrow 0+ \sin x \ln 2x\)

\(\displaystyle \sin(0+) \ln 2(0+)\)

\(\displaystyle (0) \ln (0)\)

\(\displaystyle (0) -\infty\) - Indeterminate Use L'Hopital's

\(\displaystyle \sin x \ln 2x\) - Rewrite

\(\displaystyle \dfrac{\ln 2x}{\dfrac{1}{\sin x}}\)

\(\displaystyle \dfrac{\ln 2x}{\csc x}\)

\(\displaystyle \dfrac{\dfrac{d}{dx} \ln 2x}{\dfrac{d}{dx} \csc x}\)

\(\displaystyle \dfrac{\dfrac{2}{x}}{-\csc x \cot x}\)

\(\displaystyle \dfrac{\dfrac{2}{0+}}{-\csc (0+) \cot (0+)}\)

\(\displaystyle \dfrac{\infty}{?}\)

 

[DrPhil]

\(\displaystyle \displaystyle \lim_{x \to 0+}[\sin x \ln 2x]\)

......\(\displaystyle \displaystyle = \lim_{x \to 0+}\left[\dfrac{\ln 2x}{\csc x}\right] \)

......\(\displaystyle \displaystyle = \lim_{x \to 0+}\left[\dfrac{\frac{d}{dx}( \ln 2x)}{\frac{d}{dx} (\csc x)}\right]\)

......\(\displaystyle \displaystyle = \lim_{x \to 0+}\left[\dfrac{2/2x}{-\csc x \cot x}\right]\)

............\(\displaystyle \displaystyle= -\lim_{x \to 0+} \left[\dfrac{\sin^2x}{x\ \cos x}\right] \)

If you apply l'Hospital and the result is once more indefinite, do it again!

 

[fcabanski]

It's the same thing for that other problem with e raised to the x plus x all to a power. You have to apply l'Hospital's Rule two or three times in that one.

Part of learning how to solve problems is trying things. You have a number of tools. Try them. Run into a wall. Try another tool.

............\(\displaystyle \displaystyle= -\lim_{x \to 0+} \left[\dfrac{\sin^2x}{x\ \cos x}\right] \)

If you apply l'Hospital and the result is once more indefinite, do it again!