Bài 101143

Question

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$a)$ $\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \tan 2x.\tan\left( {\frac{\pi }{4} - x} \right)$
$b)$ $\mathop {\lim }\limits_{x \to 0} \frac{3^{x^2} - \cos x}{x^2}$

hoàng anh thọ · Answer 1 · 4/25/2012 5:01:09 PM

$a)$ Đặt $t = \frac{\pi }{4} - x \Rightarrow x = \frac{\pi }{4} - t$. Tính $f(x)$ theo $t$ ta có: $f\left( x \right) = tg2x.tg\left( {\frac{\pi }{4} - x} \right) = tg\left( {\frac{\pi }{4} - 2t} \right).tgt = c{\rm{otg2t}}{\rm{.tgt = }}\frac{{c{\rm{os}}2t.\sin t}}{{\sin 2t.\cos t}} = \frac{{c{\rm{os}}2t}}{{2{{\cos }^2}t}}$
$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} f\left( x \right) = \mathop {\lim }\limits_{t \to 0} \frac{{c{\rm{os}}2t}}{{2{{\cos }^2}t}} = \frac{1}{2}$

$b)$ $f\left( x \right) = \frac{{{3^{{x^2}}} - \cos x}}{{{x^2}}} = \frac{{{3^{{x^2}}} - 1}}{{{x^2}}} + \frac{{1 - \cos x}}{{{x^2}}} = \frac{{{e^{{x^2}\ln 3}} - 1}}{{{x^2}.\ln 3}}.\ln 3 + \frac{{{{\sin }^2}x}}{{{x^2}.\left( {1 + \cos x} \right)}}$
$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}\ln 3}} - 1}}{{{x^2}.\ln 3}}.\ln 3 + \mathop {\lim }\limits_{x \to 0} {\left( {\frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{x}} \right)^2}.\frac{1}{{1 + \cos x}} = \frac{1}{2} + \ln 3$

Bài 101143

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Bài 101143

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