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l=[tex]\lim_{x \to \0} x^{sin x}[/tex]

Sagot :

LammettHash

By applying the exponential and logarithmic functions, we have

[tex]x^{\sin(x)} = \exp \left(\ln \left(x^{\sin(x)}\right)\right)[/tex]

Then in the limit,

[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \lim_{x\to0} \exp \left(\ln \left(x^{\sin(x)}\right)\right)[/tex]

[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \ln \left(x^{\sin(x)}\right)\right)[/tex]

[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \sin(x) \ln(x) \right)[/tex]

[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\ln(x)}{\csc(x)} \right)[/tex]

As x approaches 0 (from the right), both ln(x) and csc(x) approach infinity (ignoring sign). Applying L'Hopitâl's rule gives

[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp \left( \lim_{x\to0} \frac{\frac1x}{-\csc(x)\cot(x)} \right) = \exp \left( -\lim_{x\to0} \frac{\sin^2(x)}{x\cos(x)} \right)[/tex]

Recall that

[tex]\displaystyle \lim_{x\to0} \frac{\sin(x)}{x} = 1[/tex]

Then

[tex]\displaystyle \lim_{x\to0} \frac{\sin^2(x)}{x\cos(x)} = \lim_{x\to0} \frac{\sin(x)}{\cos(x)} = \lim_{x\to0} \tan(x) = 0[/tex]

So, our limit is

[tex]\displaystyle \lim_{x\to0} x^{\sin(x)} = \exp(0) = \boxed{1}[/tex]

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