Bài 107641

Question

Tìm các giới hạn sau:
a) $\mathop {\lim }\limits_{x \to 0} \frac{\\ln (2008x+1)}{x} $
b) $\mathop {\lim }\limits_{x \to 0} \frac{\\ln (5x+1)}{\sin 2x} $
c) $\mathop {\lim }\limits_{x \to 0} \frac{\\ln (2x+1)-\ln (3x+1)}{x} $
d) $\mathop {\lim }\limits_{x \to 0}\frac{e^x-1}{\sqrt{x+1}-1 }. $

Thu Hằng · Answer 1 · 6/11/2012 9:20:49 AM

Để tính các giới hạn này, ta sử dụng các kết quả sau :
$\mathop {\lim }\limits_{t \to 0}\frac{\sin t }{t} =1$ và $\mathop {\lim }\limits_{t \to 0}\frac{\ln (1+t)}{t} =1$ và $\mathop {\lim }\limits_{t \to 0}\frac{e^t-1}{t}=1 $

a) $\mathop {\lim }\limits_{x \to 0} \frac{\ln (2008x+1)}{x} =2008.\mathop {\lim }\limits_{x \to 0}\frac{\ln (2008x+1)}{2008x}=2008.1=2008 $

b) $\mathop {\lim }\limits_{x \to 0} \frac{\ln (5x+1)}{\sin 2x}=\frac{5}{2}. \mathop {\lim }\limits_{x \to 0}\frac{\frac{\ln (5x+1)}{5x} }{\frac{\sin 2x}{2x} }=\frac{5}{2}.\frac{\mathop {\lim }\limits_{x \to 0}\frac{\ln (5x+1)}{5x} }{\mathop {\lim }\limits_{x \to 0}\frac{\sin 2x}{2x} }=\frac{5}{2}. \frac{1}{1}=\frac{5}{2} $

c) $\mathop {\lim }\limits_{x \to 0} \frac{\ln (2x+1)-\ln (3x+1)}{x}=2\mathop {\lim }\limits_{x \to 0} \frac{\ln (2x+1)}{2x}-3\mathop {\lim }\limits_{x \to 0}\frac{\ln (3x+1)}{3x}=2-3=-1 $

d) $\mathop {\lim }\limits_{x \to 0} \frac{e^x-1}{\sqrt{x+1}-1 }=\mathop {\lim }\limits_{x \to 0}\frac{e^x-1}{x}.\mathop {\lim }\limits_{x \to 0}(\sqrt{x+1}+1)=1.2=2.$

Bài 107641

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Bài 107641

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