Find lim _(xarrow 0)(cosx)^(1)/(x^(2))

To find the limit \(\lim_{x \to 0} (\cos x)^{\frac{1}{x^2}}\), we can use the property of limits involving exponentials and trigonometric functions. Let's rewrite the expression in a more manageable form.<br /><br />First, observe that:<br />\[<br />(\cos x)^{\frac{1}{x^2}} = e^{\frac{1}{x^2} \ln (\cos x)}<br />\]<br /><br />Now, we need to find the limit of the exponent:<br />\[<br />\lim_{x \to 0} \frac{1}{x^2} \ln (\cos x)<br />\]<br /><br />As \(x \to 0\), \(\cos x \to 1\). We can use the Taylor series expansion for \(\cos x\) around \(x = 0\):<br />\[<br />\cos x \approx 1 - \frac{x^2}{2} + O(x^4)<br />\]<br /><br />Thus,<br />\[<br />\ln (\cos x) \approx \ln \left(1 - \frac{x^2}{2}\right) \approx -\frac{x^2}{2}<br />\]<br /><br />Therefore,<br />\[<br />\frac{1}{x^2} \ln (\cos x) \approx \frac{1}{x^2} \left(-\frac{x^2}{2}\right) = -\frac{1}{2}<br />\]<br /><br />So, the limit becomes:<br />\[<br />\lim_{x \to 0} e^{\frac{1}{x^2} \ln (\cos x)} = e^{-\frac{1}{2}}<br />\]<br /><br />Thus, the final answer is:<br />\[<br />\lim_{x \to 0} (\cos x)^{\frac{1}{x^2}} = e^{-\frac{1}{2}}<br />\]