Without L'Hôpital's

Calculus Level 2

Can you solve for the following limit without using L'Hôpital's rule ?

lim x 0 + x sin x \large \lim_{x\rightarrow0^+} x^{\sin x}


The answer is 1.

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4 solutions

ChengYiin Ong
Mar 14, 2020

I'll present the L'Hopital method: let L = lim x 0 x sin x L=\lim_ {x\to0} x^{\sin x} taking natural log on both sides, ln L = lim x 0 sin x ln x \ln L=\lim_{x\to0} \sin x \ln x = lim x 0 ln x csc x =\lim_{x\to0} \frac{\ln x}{\csc x} = lim x 0 d d x ln x d d x csc x =\lim_{x\to0} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} \csc x} = lim x 0 1 x cot x csc x =\lim_{x\to0} \frac{\frac{1}{x}}{-\cot x \csc x} = lim x 0 sin 2 x x cos x =\lim_{x\to0} -\frac{\sin ^2 x}{x\cos x } = lim x 0 sin x x tan x =\lim_{x\to0} -\frac{\sin x}{x}\tan x = 0 =0
Thus, L = e 0 = 1 L=e^0=1

You should let the limit approach 0 from above (as in x 0 + ) x \to 0^+) , since x sin x x^{\sin x} is not defined for x < 0 x < 0 .

Jon Haussmann - 1 year, 2 months ago

Since sin x x \sin x\leq x , therefore sin x \sin x approaches zero earlier than x x . So in the limit when x x tends to zero, x sin x x^{\sin x} becomes x 0 x^0 when x x is still not zero. Hence the required limit is 1 \boxed 1 .

Théo Leblanc
Apr 1, 2020

Taylor expansions are much more powerful and versatile than l'Hôpital's rule, use them !

x sin ( x ) = e ln ( x ) sin ( x ) = e ln ( x ) ( x + o ( x ) ) x 0 e 0 = 1 \begin{aligned} x^{\sin(x)} & = e^{\ln(x)\sin(x)}\\ &= e^{\ln(x)(x+o(x))} \underset{x\to 0}{\longrightarrow} e^0 =1 \end{aligned}

Note: x ln ( x ) x 0 0 x\ln(x) \underset{x\to 0}{\longrightarrow} 0

Nikola Alfredi
Mar 15, 2020

We can write lim x 0 x sin x \displaystyle \lim_{x \rightarrow 0} \ x^{\sin x} \ \ \ \ as lim x 0 e sin x ln x \ \ \ \ \ \ \displaystyle \lim_{x \rightarrow 0} \ e^{\sin x \cdot \ln x} .

As ln ( x ) = n = 1 ( 1 ) n n ( x 1 ) n \displaystyle \ln (x)= - \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}(x-1)^n \ \ This suggests, as you put x = 0 x = 0 , ln x = c c < 0 \ln x = c \ \ \ | c < 0 is a constant. But sin ( 0 ) = 0 \displaystyle \sin(0) = 0 .

Thus, lim x 0 e sin x ln x e 0 = 1 \displaystyle \lim_{x \rightarrow 0} e^{\sin x \ln x} \implies e^0 = 1 \ \ \ which is the answer.

ln ( 0 ) = \ln(0)=-\infty ... Not c c !

Théo Leblanc - 1 year, 2 months ago

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