\(lim_{x->1^-}=\dfrac{2x+1}{x-1}\)
\(lim_{x->6}=\dfrac{\left(5x-4\right)\sqrt{2x-3}+x-84}{x-6}\)
Những câu hỏi liên quan
Tính \(lim_{x\rightarrow-1}\dfrac{\sqrt{4x+5}-2x-3}{\left(x+1\right)^2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}-2x-3}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{4x+5-\left(2x+3\right)^2}{\sqrt{4x+5}+2x+3}\cdot\dfrac{1}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{4x+5-4x^2-12x-9}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4x^2-8x-4}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4\left(x^2+2x+1\right)}{\left(x+1\right)^2\cdot\left(\sqrt{4x+5}+2x+3\right)}\right)\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{-4}{\sqrt{4x+5}+2x+3}\)
\(=\dfrac{-4}{\sqrt{-4+5}-2+3}=\dfrac{-4}{1+1}=-\dfrac{4}{2}=-2\)
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cho \(lim_{x->1}\dfrac{f\left(x\right)-10}{x-1}=5\) tính giới hạn \(lim_{x->1}\dfrac{f\left(x\right)-10}{\left(\sqrt{x}-1\right)\left(\sqrt[]{4f\left(x\right)+9}+3\right)}\) bằng bao nhiêu ?
Chọn \(f\left(x\right)=5x+5\)
Khi đó: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)
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\(lim_{x->a}\left[\dfrac{1}{\left(x-a\right)^2}\left(x^2-8x+10+\dfrac{81}{x+2\sqrt{x-1}}-2\sqrt{x-1}\right)\right]=\dfrac{21}{16}\)
\(lim_{x->b}\left[\dfrac{4}{\left(x-b\right)^2}\left(x^2-x+2-2\sqrt{x}\right)\right]=c\)
với a,b,c là các số thực. Tìm a,b,c
\(lim_{x->0}\frac{x.sin2x}{1-cos2x}\)
\(lim_{x->0}\frac{\sqrt{1-x}-1}{x}\)
\(lim_{x->0-}\frac{1}{x}\left(\frac{1}{x+1}-1\right)\)
\(lim_{x->0-}\frac{2x+\sqrt{-x}}{5x-\sqrt{-x}}\)
Làm biếng viết đủ, bạn cứ tự hiểu là giới hạn khi x tiến tới gì gì đó nhé
a/ \(lim\frac{2x.sinx.cosx}{2sin^2x}=lim\frac{cosx}{\left(\frac{sinx}{x}\right)}=1\)
b/ \(lim\frac{-x}{x\left(\sqrt{1-x}+1\right)}=lim\frac{-1}{\sqrt{1-x}+1}=-\frac{1}{2}\)
c/ \(=lim\frac{1}{x}\left(\frac{x}{x+1}\right)=lim\frac{1}{x+1}=1\)
d/ \(lim\frac{\sqrt{-x}\left(2\sqrt{-x}+1\right)}{\sqrt{-x}\left(5\sqrt{-x}-1\right)}=lim\frac{2\sqrt{-x}+1}{5\sqrt{-x}-1}=\frac{1}{-1}=-1\)
\(lim_{x\rightarrow\left(-2\right)^+}\dfrac{\sqrt{8+2x}-2}{\sqrt{x+2}}\)
\(\lim\limits_{x\rightarrow\left(-2\right)^+}\dfrac{\sqrt{8+2x}-2}{\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2x+8-4}{\left(\sqrt{2x+8}+2\right)\cdot\sqrt{x+2}}\)
\(=\lim\limits_{x\rightarrow-2^+}\dfrac{2\cdot\sqrt{x+2}}{\sqrt{2x+8}+2}=\dfrac{2\cdot\sqrt{-2+2}}{\sqrt{2\cdot\left(-2\right)+8}+2}\)
=0
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\(lim_{x->1}\frac{\sqrt{6-2x}-\sqrt{x^2+3}}{\left(x-1\right)^2}\)
Tính: \(lim_{x\rightarrow-2}\dfrac{2x+1}{\left(x+2\right)^2}\)
Lời giải:
$x\to -2$ thì $2x+1\to -3<0$
$x\to -2$ thì $(x+2)^2\to 0$
$\Rightarrow \lim\limits_{x\to -2}\frac{2x+1}{(x+2)^2}=-\infty$
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\(lim_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\left(\sqrt{\dfrac{3x}{x^2-1}}\right)\)
\(\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\cdot\sqrt{\dfrac{3x}{x^2-1}}\)
\(=\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^2-x+1\right)\left(x+1\right)\cdot\dfrac{\sqrt{3x}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow\left(-1\right)^-}\sqrt{x+1}\cdot\left(x^2-x+1\right)\cdot\sqrt{\dfrac{3x}{x-1}}\)
\(=\sqrt{-1+1}\left[\left(-1\right)^2-\left(-1\right)+1\right]\cdot\sqrt{\dfrac{3\left(-1\right)}{-1-2}}\)
=0
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7 tháng 12 2023 lúc 21:15
Tìm các giới hạn sau:
a) \(lim_{x\rightarrow0}\dfrac{tan3x}{sin5x}\)
b) \(lim_{x\rightarrow0}\dfrac{cos2x-1}{sin^23x}\)
c) \(lim_{x\rightarrow1}\dfrac{x^2-4x+3}{sin\left(x-1\right)}\)