Tính lim x → 1 x 2 − a + 2 x + a + 1 x 3 − 1
A. 2 − a 3
B. − 2 − a 3
C. − a 3
D. a 3
Những câu hỏi liên quan
4. Tính giới hạn \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-x-1}{2x^2-x}_{ }\)
5. Tính giới hạn:
a) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}_{ }\)
b) \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}_{ }\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)
a. \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\lim\limits_{x\rightarrow2}\dfrac{1}{x+2}=\dfrac{1}{4}\)
b. \(\lim\limits_{x\rightarrow3^-}\dfrac{x+3}{x-3}=\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}\)
Do \(\lim\limits_{x\rightarrow3^-}\left(-x-3\right)=-6< 0\)
\(\lim\limits_{x\rightarrow3^-}\left(3-x\right)=0\) và \(3-x>0;\forall x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)
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tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x+1}{x^2+x+1}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{3x+1}{3x^2-x+5}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{3x+5}{\sqrt{x^2+x}}\)
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{-5x+1}{\sqrt{3x^2+1}}\)
`a)lim_{x->+oo}[x+1]/[x^2+x+1]`
`=lim_{x->+oo}[1/x+1/[x^2]]/[1+1/x+1/[x^2]]`
`=0`
`b)lim_{x->+oo}[3x+1]/[3x^2-x+5]`
`=lim_{x->+oo}[3/x+1/[x^2]]/[3-1/x+5/[x^2]]`
`=0`
`c)lim_{x->-oo}[3x+5]/[\sqrt{x^2+x}]`
`=lim_{x->-oo}[3+5/x]/[-\sqrt{1+1/x}]`
`=-3`
`d)lim_{x->+oo}[-5x+1]/[\sqrt{3x^2+1}]`
`=lim_{x->+oo}[-5+1/x]/[\sqrt{3+1/[x^2]}]`
`=-5/3`
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Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow1^+}\dfrac{x^3+x+1}{x-1}\)
b) \(\lim\limits_{x\rightarrow-1^+}\dfrac{3x+2}{x+1}\)
c) \(\lim\limits_{x\rightarrow2^-}\dfrac{x-15}{x-2}\)
Lời giải:
a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)
\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$
\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)
b.
\(\lim\limits_{x\to -1+}(3x+2)=-1<0\)
\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$
\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)
c.
\(\lim\limits_{x\to 2-}(x-15)=-17<0\)
\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
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Cho
lim
x
→
∞
f
x
+
2
1
.Tính
lim
...
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Cho lim x → ∞ f x + 2 = 1 .Tính lim x → ∞ f x
A. lim x → ∞ f x = 3
B. lim x → ∞ f x = -1
C. lim x → ∞ f x = -3
D. lim x → ∞ f x = 1
ai tìm ra cách sai trong 2 cái giải này giúp mình với: đề bài là tính limsqrt{x^4+x^2}-sqrt[3]{x^6+1}C1:limsqrt{x^4+x^2}-sqrt[3]{x^6+1}limleft(x^2left(sqrt{1+dfrac{1}{x^2}}right)-sqrt[3]{1+dfrac{1}{x^6}}right)lim x2(1-1)0C2:limsqrt{x^4+x^2}-sqrt[3]{x^6+1}limleft(sqrt{x^4+x^2}-x^2-sqrt[3]{x^6+1}+x^2right)
limleft(dfrac{x^2}{sqrt{x^4+x^2}+x^2}-dfrac{1}{left(sqrt[3]{x^6+1}right)^2+x^2.sqrt[3]{x^6+1}+x^4}right)lim(dfrac{1}{2}-0) dfrac{1}{2}mình không biết cách nào đúng ai chỉ cho mình với
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ai tìm ra cách sai trong 2 cái giải này giúp mình với: đề bài là tính \(lim\sqrt{x^4+x^2}-\sqrt[3]{x^6+1}\)
C1:\(lim\sqrt{x^4+x^2}-\sqrt[3]{x^6+1}=lim\left(x^2\left(\sqrt{1+\dfrac{1}{x^2}}\right)-\sqrt[3]{1+\dfrac{1}{x^6}}\right)\)=lim x2(1-1)=0
C2:\(lim\sqrt{x^4+x^2}-\sqrt[3]{x^6+1}=lim\left(\sqrt{x^4+x^2}-x^2-\sqrt[3]{x^6+1}+x^2\right)\\ \)=\(lim\left(\dfrac{x^2}{\sqrt{x^4+x^2}+x^2}-\dfrac{1}{\left(\sqrt[3]{x^6+1}\right)^2+x^2.\sqrt[3]{x^6+1}+x^4}\right)\)
=lim(\(\dfrac{1}{2}-0\))= \(\dfrac{1}{2}\)
mình không biết cách nào đúng ai chỉ cho mình với
Hiển nhiên là cách đầu sai rồi em
Khi đến \(\lim x^2\left(1-1\right)=+\infty.0\) là 1 dạng vô định khác, đâu thể kết luận nó bằng 0 được
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tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^2+x^3+5}{4x^3+1}\)
b) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]` `ĐK: 4x^3+1 ne 0`
`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`
`=1/4`
`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]` `ĐK: x ne 0;x ne 1`
`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`
`=0`
Câu `c` giống `b`.
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Cho hai hàm số fleft( x right) {x^2} - 1,gleft( x right) x + 1.a) Tính mathop {lim }limits_{x to 1} fleft( x right) và mathop {lim }limits_{x to 1} gleft( x right).b) Tính mathop {lim }limits_{x to 1} left[ {fleft( x right) + gleft( x right)} right]và so sánh mathop {lim }limits_{x to 1} fleft( x right) + mathop {lim }limits_{x to 1} gleft( x right).c) Tính mathop {lim }limits_{x to 1} left[ {fleft( x right) - gleft( x right)} right]và so sánh mathop {lim }limits_{x to 1} fleft( x right) - mat...
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Cho hai hàm số \(f\left( x \right) = {x^2} - 1,g\left( x \right) = x + 1.\)
a) Tính \(\mathop {\lim }\limits_{x \to 1} f\left( x \right)\) và \(\mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
b) Tính \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right]\)và so sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
c) Tính \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right]\)và so sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
d) Tính \(\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right]\)và so sánh \(\mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right).\)
e) Tính \(\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}}\)và so sánh \(\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\)
a) \(\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 1} \right) = \mathop {\lim }\limits_{x \to 1} {x^2} - \mathop {\lim }\limits_{x \to 1} 1 = {1^2} - 1 = 0\)
\(\mathop {\lim }\limits_{x \to 1} g\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = \mathop {\lim }\limits_{x \to 1} x + \mathop {\lim }\limits_{x \to 1} 1 = 1 + 1 = 2\)
b) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} + x} \right) = {1^2} + 1 = 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 + 2 = 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) + \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
c) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - x - 2} \right) = {1^2} - 1 - 2 = - 2\\\mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0 - 2 = - 2\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right) - g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right) - \mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
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d) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left[ {\left( {{x^2} - 1} \right)\left( {x + 1} \right)} \right] = \mathop {\lim }\limits_{x \to 1} \left( {{x^3} + {x^2} - x - 1} \right) = {1^3} + {1^2} - 1 - 1 = 0\\\mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right) = 0.2 = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \left[ {f\left( x \right).g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to 1} f\left( x \right).\mathop {\lim }\limits_{x \to 1} g\left( x \right).\end{array}\)
e) \(\begin{array}{l}\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x - 1} \right) = 1 - 1 = 0\\\frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}} = \frac{0}{2} = 0\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to 1} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to 1} g\left( x \right)}}.\end{array}\)
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tính các giới hạn sau:
a. \(lim\dfrac{\sqrt{x+1}-x+1}{x^2-5x+6}\)
x->3
b. \(lim\left|x^3-3x\right|\)
x->-2
Câu a.
\(^{lim}_{x\rightarrow3}\dfrac{\sqrt{x+1}-x+1}{x^2-5x+6}\)
Nhân liên hợp ta đc:
\(^{lim}_{x\rightarrow3}\dfrac{x+1-\left(x-1\right)^2}{(x^2-5x+6)\cdot\left(\sqrt{x+1}+x-1\right)}\)
\(=^{lim}_{x\rightarrow3}\dfrac{-x^2+3x}{\left(x-3\right)\left(x-2\right)\left(\sqrt{x+1}+x-1\right)}\)
\(=^{lim}_{x\rightarrow3}\dfrac{-x}{\left(x-2\right)\cdot\left(\sqrt{x+1}+x-1\right)}\)
\(=\dfrac{-3}{\left(3-2\right)\cdot\left(\sqrt{3+1}+3-1\right)}=-\dfrac{3}{4}\)
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Câu b.
\(^{lim}_{x\rightarrow-2}\left|x^3-3x\right|\)
\(=\left|\left(-2\right)^3-3\cdot\left(-2\right)\right|=\left|-2\right|=2\)
Câu này đơn giản chỉ thay số thôi nhé, nó ở dạng đa thức nữa!
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tính giới hạn
a) \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)
b) \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
a: \(\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{\sqrt{3x+1}-2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{\dfrac{3x+1-4}{\sqrt{3x+1}+2}}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)\cdot\left(\sqrt{3x+1}+2\right)}{3\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}{3}\)
\(=\dfrac{\left(1+1\right)\left(\sqrt{3+1}+2\right)}{2}=\dfrac{2\cdot4}{3}=\dfrac{8}{3}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-2x}{\sqrt{x+2}-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{\dfrac{x+2-4}{\sqrt{x+2}+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)\cdot\left(\sqrt{x+2}+2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}x\left(\sqrt{x+2}+2\right)\)
\(=2\cdot\left(\sqrt{2+2}+2\right)\)
\(=2\cdot4=8\)
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tính giới hạn
a) \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
b) \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
c) \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
a: \(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{2x+8}-4}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2x+8-16}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{\sqrt{2x+8}+4}\cdot\dfrac{1}{x-4}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{2}{\sqrt{2x+8}+4}=\dfrac{2}{\sqrt{2\cdot4+8}+4}\)
\(=\dfrac{2}{\sqrt{8+8}+4}=\dfrac{2}{4+4}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{\sqrt{4x+1}-3}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{\dfrac{4x+1-9}{\sqrt{4x+1}+3}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2\right)}{4\left(x-2\right)}\cdot\left(\sqrt{4x+1}+3\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x+2\right)\left(\sqrt{4x+1}+3\right)}{4}\)
\(=\dfrac{\left(2+2\right)\left(\sqrt{4\cdot2+1}+3\right)}{4}=\sqrt{9}+3=6\)
c: \(\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{\dfrac{4-x-2}{2+\sqrt{x+2}}}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x-2}{2-x}\cdot\left(\sqrt{x+2}+2\right)\)
\(=\lim\limits_{x\rightarrow2}\left(-\sqrt{x+2}-2\right)\)
\(=-\sqrt{2+2}-2=-2-2=-4\)
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