Bài viết đã được chỉnh sửa nội dung bởi mileycyrus: 01-07-2011 - 23:15
tìm giới hạn
Bắt đầu bởi mileycyrus, 01-07-2011 - 23:14
#1
Đã gửi 01-07-2011 - 23:14
$ \lim_{x\to 1}\dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1} $
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#2
Đã gửi 02-07-2011 - 08:13
$\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}=\left(\dfrac{0}{0}\right)$ form
Use D.L Hospital rule:
Differentiate numerator and Denominator seperately:
$\displaystyle \lim_{x\rightarrow 1}\dfrac{\dfrac{2}{2.\sqrt{2x-1}}+2x-3}{\dfrac{1}{3.\sqrt[3]{(x-2)^2}}+2x-1}=\left(\dfrac{0}{0}\right)$ form
Again Differentiate numerator and Denominator seperately:
$\displaystyle \lim_{x\rightarrow 1}\dfrac{-\dfrac{1}{2}.(2x-1)^{-\dfrac{3}{2}}\times 2+2}{\dfrac{1}{3}.-\dfrac{2}{3}\times (x-2)^{-\dfrac{5}{3}}+2}$
$\displaystyle = \dfrac{9}{20}$
Use D.L Hospital rule:
Differentiate numerator and Denominator seperately:
$\displaystyle \lim_{x\rightarrow 1}\dfrac{\dfrac{2}{2.\sqrt{2x-1}}+2x-3}{\dfrac{1}{3.\sqrt[3]{(x-2)^2}}+2x-1}=\left(\dfrac{0}{0}\right)$ form
Again Differentiate numerator and Denominator seperately:
$\displaystyle \lim_{x\rightarrow 1}\dfrac{-\dfrac{1}{2}.(2x-1)^{-\dfrac{3}{2}}\times 2+2}{\dfrac{1}{3}.-\dfrac{2}{3}\times (x-2)^{-\dfrac{5}{3}}+2}$
$\displaystyle = \dfrac{9}{20}$
#3
Đã gửi 02-07-2011 - 18:46
Another solution:
$\lim_{x \to 1}\dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}=\lim_{x \to 1}\dfrac{\dfrac{2}{\sqrt{2x-1}+1}+x-2}{\dfrac{1}{\sqrt[3]{(x-2)^2}+\sqrt[3]{x-2}+1}+x}$
$=\lim_{x \to 1}\dfrac{1-\dfrac{2}{(\sqrt{2x-1}+1)^2}}{\dfrac{\dfrac{x-3}{1+\sqrt[3]{(x-2)^4}+\sqrt[3]{(x-2)^2}}+\dfrac{1}{1+\sqrt[3]{x-2}+\sqrt[3]{(x-2)^2}}}{\sqrt[3]{(x-2)^2}+\sqrt[3]{x-2}+1}+1}$.
And we only have tp replace $x=1$ to the above expression.Done.
P/s:@stuart clark: I think my solution is easier to understand than yours.
$\lim_{x \to 1}\dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}=\lim_{x \to 1}\dfrac{\dfrac{2}{\sqrt{2x-1}+1}+x-2}{\dfrac{1}{\sqrt[3]{(x-2)^2}+\sqrt[3]{x-2}+1}+x}$
$=\lim_{x \to 1}\dfrac{1-\dfrac{2}{(\sqrt{2x-1}+1)^2}}{\dfrac{\dfrac{x-3}{1+\sqrt[3]{(x-2)^4}+\sqrt[3]{(x-2)^2}}+\dfrac{1}{1+\sqrt[3]{x-2}+\sqrt[3]{(x-2)^2}}}{\sqrt[3]{(x-2)^2}+\sqrt[3]{x-2}+1}+1}$.
And we only have tp replace $x=1$ to the above expression.Done.
P/s:@stuart clark: I think my solution is easier to understand than yours.
Bài viết đã được chỉnh sửa nội dung bởi dark templar: 02-07-2011 - 18:50
"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.
#4
Đã gửi 03-07-2011 - 14:08
1)$ \lim_{x\to 0}\dfrac{ \sqrt[3]{1+x^2}- \sqrt[4]{1-2x}}{x+x^2} $
2)$ \lim_{x\to +00}\dfrac{ \sqrt[]{x}- \sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{2x+1}} $
2)$ \lim_{x\to +00}\dfrac{ \sqrt[]{x}- \sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{2x+1}} $
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#5
Đã gửi 04-07-2011 - 10:03
Bài 1:1)$ \lim_{x\to 0}\dfrac{ \sqrt[3]{1+x^2}- \sqrt[4]{1-2x}}{x+x^2} $
2)$ \lim_{x\to +00}\dfrac{ \sqrt[]{x}- \sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{2x+1}} $
$\lim_{x \to 0}\dfrac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}=\lim_{x \to 0}\dfrac{(\sqrt[3]{1+x^2}-1)-(\sqrt[4]{1-2x}-1)}{x.(1+x)}$
$=\lim_{x \to 0}\dfrac{\dfrac{x}{\sqrt[3]{(1+x^2)^2}+\sqrt[3]{1+x^2}+1}+\dfrac{2}{(\sqrt[4]{1-2x}+1)(\sqrt{1-2x}+1)}}{1+x}$
Đến đây thì chỉ cần thế $x=0$ vào là được .Thực ra bài này cũng có thể xài quy tắc D.L.Hospital giống như bạn stuart clark đã làm,nhưng do đó là chương trình ĐH nên mình không nêu ra ở đây.
Bài 2:
Có phải là cho $x \to + \infty$ không? Bởi nếu cho $x \to 0$ thì quá dễ r�ồi. Nếu cho $x \to + \infty$ thì làm như sau:
$\lim_{x \to + \infty}\dfrac{\sqrt{x}-\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{1+2x}}=\lim_{x \to + \infty}\dfrac{1-\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{2+\dfrac{1}{x}}}$
Có $\lim_{x \to + \infty}\dfrac{1}{\sqrt[4]{x}}=\lim_{x \to + \infty}\dfrac{1}{\sqrt[6]{x}}=\lim_{x \to + \infty}\dfrac{1}{x}=0$ nên:
$\lim_{x \to + \infty}\dfrac{1-\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{2+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{2}}$.
Xong.
Bài viết đã được chỉnh sửa nội dung bởi dark templar: 04-07-2011 - 10:05
"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.
#6
Đã gửi 04-07-2011 - 11:08
Tìm các giới hạn:
1)$ \lim_{x\to 0}\dfrac{1-\sqrt{2x+1}+sinx}{\sqrt{3x+4}-2-x} $
2)$ \mathop {\lim }\limits_{x \to 0} \dfrac{{cos[(\dfrac{\pi }{2}).cosx]}}{{si{n^2}(x/2)}} $
1)$ \lim_{x\to 0}\dfrac{1-\sqrt{2x+1}+sinx}{\sqrt{3x+4}-2-x} $
2)$ \mathop {\lim }\limits_{x \to 0} \dfrac{{cos[(\dfrac{\pi }{2}).cosx]}}{{si{n^2}(x/2)}} $
Bài viết đã được chỉnh sửa nội dung bởi mileycyrus: 04-07-2011 - 11:09
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#7
Đã gửi 10-07-2011 - 16:48
$(1)\;\;\; \displaystyle \lim_{x\rightarrow 0}\dfrac{1-\sqrt{2x+1}+\sin x}{\sqrt{3x+4} -2 -x} = \dfrac{0}{0}$ form
apply $D.L.H$ rule
$\displaystyle \lim_{x\rightarrow 0}\dfrac{-\dfrac{2}{2.\sqrt{2x+1}}+\cos x}{\dfrac{3}{3.\sqrt{3x+4}}-1} = 0$
$(2)\;\;\; \displaystyle \lim_{x\rightarrow 0}\dfrac{\cos \left(\dfrac{\pi}{2}.\cos x\right)}{\sin^2 \dfrac{x}{2}} = \dfrac{0}{0}$ form
$\displaystyle \lim_{x \rightarrow 0}\dfrac{\cos \left(\dfrac{\pi}{2}.\cos x\right)}{\dfrac{\sin^2 \dfrac{x}{2}}{\dfrac{x^2}{4}}.\dfrac{x^2}{4}}=\lim_{x \rightarrow 0}\dfrac{4.\cos \left(\dfrac{\pi}{2}.\cos x\right)}{x^2}=\dfrac{0}{0}$ form
apply $D.L.H$ rule
$\displaystyle \lim_{x \rightarrow 0}\dfrac{-4.\sin\left(\dfrac{\pi}{2}.\cos x\right)\times -\dfrac{\pi}{2}\sin x}{2x}=\pi$
apply $D.L.H$ rule
$\displaystyle \lim_{x\rightarrow 0}\dfrac{-\dfrac{2}{2.\sqrt{2x+1}}+\cos x}{\dfrac{3}{3.\sqrt{3x+4}}-1} = 0$
$(2)\;\;\; \displaystyle \lim_{x\rightarrow 0}\dfrac{\cos \left(\dfrac{\pi}{2}.\cos x\right)}{\sin^2 \dfrac{x}{2}} = \dfrac{0}{0}$ form
$\displaystyle \lim_{x \rightarrow 0}\dfrac{\cos \left(\dfrac{\pi}{2}.\cos x\right)}{\dfrac{\sin^2 \dfrac{x}{2}}{\dfrac{x^2}{4}}.\dfrac{x^2}{4}}=\lim_{x \rightarrow 0}\dfrac{4.\cos \left(\dfrac{\pi}{2}.\cos x\right)}{x^2}=\dfrac{0}{0}$ form
apply $D.L.H$ rule
$\displaystyle \lim_{x \rightarrow 0}\dfrac{-4.\sin\left(\dfrac{\pi}{2}.\cos x\right)\times -\dfrac{\pi}{2}\sin x}{2x}=\pi$
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