Limits of powers
Task number: 3031
Determine the following limits:
Variant 1
\( \displaystyle \lim_{x\to 0^+} x^x \)
Resolution
\(x^x=e^{x\ln x}=e^{\frac{\ln x}{x^{-1}}}\)
Because \( \displaystyle \lim_{x\to 0^+}\frac{(\ln x)'}{(x^{-1})'}= \lim_{x\to 0^+}\frac{\frac1x}{-x^{-2}}= \lim_{x\to 0^+}-x=0 \)
and the denominator is non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to 0^+} x^x=e^0=1 \)
Result
The limit is 1.
Variant 2
\( \displaystyle \lim_{x\to 0^+}\left(\ln\frac1x\right)^x \)
Resolution
\(\left(\ln\frac1x\right)^x=(-\ln x)^x=-e^{x\ln(-\ln x)}=e^{\frac{\ln(-\ln x)}{x^{-1}}}\)
Because \( \displaystyle \lim_{x\to 0^+}\frac{(\ln(-\ln x))'}{(x^{-1})'}= \lim_{x\to 0^+}\frac{-\frac1{x\ln x}}{-x^{-2}}= \lim_{x\to 0^+}\frac{x}{\ln x}=0 \)
and the denominator is non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to 0^+} x^x=e^0=1 \)
Variant 3
\( \displaystyle \lim_{x\to 0} (e^x+x)^{\frac1x} \)
Resolution
\((e^x+x)^{\frac1x}=e^{\frac{\ln (e^x+x)}{x}}\)
Because \( \displaystyle \lim_{x\to 0^+}\frac{(\ln (e^x+x))'}{(x)'}= \lim_{x\to 0^+}\frac{\frac{e^x+1}{e^x+x}}{1}= \lim_{x\to 0^+}\frac{e^x+1}{e^x+x}=2 \)
and the denominator is non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to 0^+} (e^x+x)^{\frac1x}=e^2 \)
Result
The limit is \(e^2\).
Variant 4
\( \displaystyle \lim_{x\to 0}(e^x-4x)^{\frac1x} \)
Resolution
\((e^x-4x)^{\frac1x}=e^{\frac{\ln (e^x-4x)}{x}}\)
Because \( \displaystyle \lim_{x\to 0^+}\frac{(\ln (e^x-4x))'}{(x)'}= \lim_{x\to 0^+}\frac{\frac{e^x-4}{e^x-4x}}{1}= \lim_{x\to 0^+}\frac{e^x-4}{e^x-4x}=-3 \)
and the denominator is non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to 0^+} (e^x-4x)^{\frac1x}=e^{-3} \)
Result
The limit is \(e^{-3}\).
Variant 5
\( \displaystyle \lim_{x\to \infty}\left(1-\frac2x\right)^x \)
Resolution
\(\left(1-\frac2x\right)^x=e^{x\ln\left(1-\frac2x\right)}\)
Because \( \displaystyle \lim_{x\to\infty}\frac{\left(\ln\left(1-\frac2x\right)\right)'}{(x^{-1})'}= \lim_{x\to\infty}\frac{\frac1{1-\frac2x}\cdot\frac2{x^2}}{-x^{-2}}=-2 \)
and the denominator is non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to \infty} \left(1-\frac2x\right)^x=e^{-2} \)
Result
The limit is \(e^{-2}\).
Variant 6
\( \displaystyle \lim_{x\to 0}(\cos x)^{\cot^2 x} \)
Resolution
\(\displaystyle (\cos x)^{\cot^2 x}=(\cos x)^{\frac{\cos^2 x}{1-\cos^2 x}} \)
Using the theorem about the limit of a composite function, we determine the limit
\( \displaystyle \lim_{x\to 1}x^{\frac{x^2}{1-x^2}}= \lim_{x\to 1}e^{\frac{\ln x\cdot x^2}{1-x^2}} \)
Because \( \displaystyle \lim_{x\to 1}\frac{\left(\ln x\cdot x^2 \right)'}{(1-x^2)'}= \lim_{x\to 1}\frac{\frac{x^2}{x}+2x\ln x}{-2x}= -\frac12-\lim_{x\to 1}\ln x=\frac12 \)
and the denominator is non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to 0}(\cos x)^{\cot^2 x}= e^{-\frac 12} \)
Result
The limit is \(e^{-\frac 12}\).
Variant 7
\( \displaystyle \lim_{x\to 0} \left(\frac{\sin x}x\right)^{\frac1{x^2}} \)
Hint
For a fraction of the type \(\frac 00\) you can repeatedly apply l'Hospital's rule.
Resolution
\(\displaystyle \left(\frac{\sin x}x\right)^{\frac1{x^2}} =e^{\frac{\ln \sin x -\ln x}{x^2}}\)
Because: \( \displaystyle \lim_{x\to 0}\frac{(\ln \sin x -\ln x)'}{(x^{2})'}= \lim_{x\to 0}\frac{\cot x -\frac 1x}{2x} \)
\( \displaystyle \lim_{x\to 0}\frac{\left(\cot x -\frac 1x\right)'}{(2x)'}= \lim_{x\to 0}\frac{\left(\cot x -\frac 1x\right)'}{2}= \lim_{x\to 0}\frac1{2x^2}-\frac1{2\sin^2 x}= \lim_{x\to 0}\frac{\sin^2 x-x^2}{2x^2\sin^2 x} \),
\( \displaystyle \lim_{x\to 0}\frac{\left(\sin^2 x-x^2\right)'}{\left(2x^2\sin^2 x\right)'}= \lim_{x\to 0}\frac{2\sin x\cos x-2x}{4x^2\sin x\cos x+4x\sin^2 x}= \lim_{x\to 0}\frac{\sin x\cos x-x}{2x^2\sin x\cos x+2x\sin^2 x} \),
\( \displaystyle \lim_{x\to 0}\frac{(\sin x\cos x-x)'}{(2x^2\sin x\cos x+2x\sin^2 x)'}= \lim_{x\to 0}\frac{-\sin^2 x+\cos^2 x-1}{4x\sin x\cos x+2x^2\cos^2 x-2x^2\sin^2x+2\sin^2 x+4x\sin x\cos x}= \)
\[ =\lim_{x\to 0}\frac{-\sin^2 x}{4x\sin x\cos x+x^2\cos^2 x-x^2\sin^2x+\sin^2 x}\,, \]\( \displaystyle \lim_{x\to 0}\frac{(-\sin^2 x)'}{(4x\sin x\cos x+x^2\cos^2 x-x^2\sin^2x+\sin^2 x)'}=\\ \lim_{x\to 0}\frac{-2\sin x\cos x}{4\sin x\cos x+4x\cos^2 x-4x\sin^2 x+2x\cos^2 x-2x^2\sin x\cos x - 2x\sin^2 x- 2x^2\sin x\cos x +2\sin x\cos x}=\\ \lim_{x\to 0}\frac{-\sin x\cos x}{3\sin x\cos x+3x\cos^2 x -3x\sin^2 x-2x^2\sin x\cos x } \),
\( \displaystyle \lim_{x\to 0}\frac{(-\sin x\cos x)'}{(3\sin x\cos x+3x\cos^2 x -3x\sin^2 x-2x^2\sin x\cos x )'}=\\ \lim_{x\to 0}\frac{-\cos^2 x+\sin^2 x}{3\cos^2 x-3 \sin^2 x+3\cos^2 x-6x\cos x\sin x -3\sin^2 x- 6x\sin x\cos x-4x\sin x\cos x-2x^2\cos^2x+2x^2\sin^2 x }=-\frac 16 \),
and the denominators are non-zero in a neighborhood of \(0\), we have
\( \displaystyle \lim_{x\to 0} \left(\frac{\sin x}x\right)^{\frac1{x^2}} =e^{-\frac16} \)
Result
The limit is \(e^{-\frac 16}\).
