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Vlada [557]
3 years ago
10

What is 25.2<1.5y+1.2

Mathematics
1 answer:
valentinak56 [21]3 years ago
7 0
25,2
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Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))
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I will make the assumption that <span>log(x)=ln(x)</span><span>.

The limit result can be proven if the base of </span><span>log(x)</span><span> is 10. 
</span>
\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x  \log x }  \\~\\  \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x)  }   \\~\\  \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x)  }  ~~ \normalsize{\text{ substituting x for sin x } } \\~\\   \large  = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log(  \lim_{x \to 0^{+}}x^x)  } = \frac{1-1}{\log(1)}   = \frac{0}{0}

We get the indeterminate form 0/0, so we have to use <span>Lhopitals rule 

</span>\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x)  } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x)  }   \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1
<span>
Therefore,

</span>\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x  \log x }  =\boxed{ -1}<span>
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