Bài 107667

Question

Tính các giới hạn sau:
a) $\mathop {\lim }\limits_{x \to + \infty} (1+\frac{a}{x} )^x, (a \neq 0) $ b) $\mathop {\lim }\limits_{x \to 0} (\frac{e^{ax}-e^{bx}}{x} ), (a \neq 0, b \neq 0)$
c) $\mathop {\lim }\limits_{x \to + \infty} (\frac{x+1}{x-1} )^{x+2}$ d) $\mathop {\lim }\limits_{x \to 0} (1 + \sin x)^ \frac{1}{x}$

Thu Hằng · Answer 1 · 6/11/2012 11:00:46 AM

Ta sẽ sử dụng (không chứng minh) kết quả sau :
$\mathop {\lim }\limits_{t \to + \infty} \left ( 1 + \frac{1}{t} \right )^t=\mathop {\lim }\limits_{m \to 0} (1 + m)^ \frac{1}{m} =e$

a) $\mathop {\lim }\limits_{x \to + \infty} \left ( 1 + \frac{a}{x} \right )^x = \mathop {\lim }\limits_{t \to + \infty} \left[ {\left ( 1 + \frac{1}{t}\right)^t } \right]^a = e^a$, với $x = at$

b) $\mathop {\lim }\limits_{x \to 0} \left ( \frac{e^{ax} - e^{bx}}{x}\right ) = \mathop {\lim }\limits_{x \to 0} \left ( \frac{(e^{ax} - 1) - (e^{bx} - 1)}{x} \right ) $
$= a \mathop {\lim }\limits_{x \to 0} \left ( \frac{e^{ax} - 1}{ax} \right ) - b \mathop {\lim }\limits_{x \to 0} \left ( \frac{e^{bx}-1}{bx} \right ) = a - b$

c) $\mathop {\lim }\limits_{x \to + \infty} \left ( \frac{x + 1}{x - 1} \right )^{x+2} = \mathop {\lim }\limits_{x \to + \infty} \left ( 1 + \frac{2}{x-1} \right )^{x + 2}$
Đặt $\frac{2}{x - 1} = \frac{1}{t} \Rightarrow x = 2t + 1$
Ta có $\mathop {\lim }\limits_{x \to + \infty} \left ( \frac{x + 1}{x - 1} \right )^{x+2} = \mathop {\lim }\limits_{t \to + \infty} \left ( 1 + \frac{1}{t} \right )^{2t+3} = \mathop {\lim }\limits_{t \to + \infty} \left[ {\left ( 1 + \frac{1}{t}\right)^t } \right]^2 . \mathop {\lim }\limits_{t \to + \infty} \left ( 1 + \frac{1}{t} \right )^3 $
$= e^2.1 = e^2$

d) $\mathop {\lim }\limits_{x \to 0} (1 + \sin x)^ \frac{1}{x} = \mathop {\lim }\limits_{x \to 0} \left[ {(1 + \sin x)^ \frac{1}{\sin x} } \right] ^ \frac{\sin x}{x} = e^1 = e$

Bài 107667

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Bài 107667

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