Limit involving x \left |x \right |

Calculus Level 3

lim x 0 x sin x \large \lim_{x \rightarrow 0}\left |x \right |^{\sin x}

Find the above limit, if it exists.

Clarification : e 2.71828 e \approx 2.71828 is the Euler's number .

1 1 0 0 e e e 2 e^2 Limit does not exist

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3 solutions

Sam Bealing
May 31, 2016

y = lim x 0 x sin x y=\lim_{x \rightarrow 0} x^{\sin{x}}

ln y = lim x 0 ( sin x ) ( ln x ) = lim x 0 ln x csc x \ln{y}=\lim_{x \rightarrow 0} (\sin{x}) (\ln{x})=\lim_{x \rightarrow 0} \dfrac{\ln{x}}{\csc{x}}

Using L'Hopital's Rule :

= lim x 0 1 x cot x csc x = lim x 0 sin 2 x x cos x = ( lim x 0 1 cos x ) ( lim x 0 sin 2 x x ) = lim x 0 sin 2 x x = lim x 0 2 sin x cos x 1 = lim x 0 sin 2 x = 0 \cdots=\lim_{x \rightarrow 0} \dfrac{\frac{1}{x}}{-\cot{x} \csc{x}}=-\lim_{x \rightarrow 0} \dfrac{\sin^2{x}}{x \cos{x}}=-\left (\lim_{x \rightarrow 0} \dfrac{1}{\cos{x}} \right) \left (\lim_{x \rightarrow 0} \dfrac{\sin^2{x}}{x} \right)=-\lim_{x \rightarrow 0} \dfrac{\sin^2{x}}{x} \\ \cdots=-\lim_{x \rightarrow 0} \dfrac{2\sin{x} \cos{x}}{1}=-\lim_{x \rightarrow 0} \sin{2x}=0

ln y = 0 y = e 0 = 1 \ln{y}=0 \Rightarrow y=e^0=\boxed{1}

Moderator note:

Simple standard approach.

Chew-Seong Cheong
May 31, 2016

The limit L = lim x 0 x sin x \displaystyle L = \lim_{x \to 0} \ |x|^{\sin x} exists if lim x 0 + x sin x = lim x 0 x sin x \displaystyle \lim_{x \to 0^+} |x|^{\sin x} = \lim_{x \to 0^-} |x|^{\sin x} . Now consider:

L + = lim x 0 + x sin x = lim x 0 + x sin x = lim x 0 + e ln x sin x Expressing e x by a more convenient form exp ( x ) = lim x 0 + exp ( ln x sin x ) Dividing up and down by sin x = lim x 0 + exp ( ln x csc x ) Since ln x and csc x as x 0 + , we can use the L’H o ˆ pital’s rule. = lim x 0 + exp ( 1 x csc x cot x ) Differentiating up and down = lim x 0 + exp ( sin x sin x x cos x ) = exp ( ( 1 ) ( 0 ) 1 ) = e 0 = 1 \begin{aligned} L_+ & = \lim_{x\to 0^+} |x|^{\sin x} \\ & = \lim_{x\to 0^+} x^{\sin x} \\ & = \lim_{x\to 0^+} \color{#3D99F6}{e}^{\ln x \sin x} \quad \quad \small \color{#3D99F6}{\text{Expressing } e^x \text{ by a more convenient form }\exp(x)} \\ & = \lim_{x\to 0^+} \color{#3D99F6}{\exp }(\ln x \sin x) \quad \quad \small \color{#3D99F6}{\text{Dividing up and down by }\sin x} \\ & = \lim_{x\to 0^+} \exp \left(\frac{\ln x}{\csc x}\right) \quad \quad \small \color{#3D99F6}{\text{Since } \ln x \to -\infty \text{ and }\csc x \to \infty \text{ as }x \to 0^+ \text{, we can use the L'Hôpital's rule.}} \\ & = \lim_{x\to 0^+} \exp \left(\frac{\frac{1}{x}}{-\csc x \cot x}\right) \quad \quad \small \color{#3D99F6}{\text{Differentiating up and down}} \\ & = \lim_{x\to 0^+} \exp \left(\frac{\color{#3D99F6}{\sin x}\sin x}{\color{#3D99F6}{x}\cos x}\right) \\ & = \exp \left(\frac{(\color{#3D99F6}{1})(0)}{1}\right) \\ & = e^0 = 1 \end{aligned}

Now, let us look at:

L = lim x 0 x sin x = lim x 0 + x sin x = lim x 0 + x sin x = lim x 0 + 1 x sin x = 1 L + = 1 \begin{aligned} L_- & = \lim_{x\to 0^-} |x|^{\sin x} \\ & = \lim_{x\to 0^\color{#D61F06}{+}} |\color{#D61F06}{-}x|^{\color{#D61F06}{-}\sin x} \\ & = \lim_{x\to 0^\color{#D61F06}{+}} x^{\color{#D61F06}{-}\sin x} \\ & = \lim_{x\to 0^\color{#D61F06}{+}} \frac 1 {x^{\sin x}} \\ & = \frac1L_+ = 1 \end{aligned}

Therefore, L = L + = L = 1 L = L_+ = L_- = \boxed{1}

Krishna Shankar
Jun 26, 2016

Easier way is to plot a graph of the given limit. See at zero, the limit is one.

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