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\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-\left(x+1\right)}{2x^2-x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{x^2+1}-\left(x+1\right)\right)\left(\sqrt{x^2+1}+x+1\right)}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2x}{x\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-2}{\left(2x-1\right)\left(\sqrt{x^2+1}+x+1\right)}\)
\(=\dfrac{-2}{\left(0-1\right)\left(\sqrt{1}+1\right)}=1\)
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 \(3-x>0;\forall x< 3\)
\(\Rightarrow\lim\limits_{x\rightarrow3^-}\dfrac{-x-3}{3-x}=-\infty\)
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
Ta có : (...) = \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)-\left[\sqrt[3]{x^2+7}-\left(x+1\right)\right]}{x^2-1}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{5-x^3-\left(x+1\right)^2}{\left(\sqrt{5-x^3}+x+1\right)\left(x^2-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-x^3-x^2-2x+4}{...}\) \(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)\left(x-1\right)}{...}\)
= \(\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+2x+4\right)}{\left(x+1\right)\left(\sqrt{5-x^3}+x+1\right)}=\dfrac{-7}{8}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\left(x+1\right)}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{x^2+7-x^3-3x^2-3x-1}{\left(x^2-1\right)\left[\sqrt[3]{\left(x+7\right)^2}+\left(x+1\right)\sqrt[3]{x^2+7}+\left(x+1\right)^2\right]}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)\left(x-1\right)}{...}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{-\left(x^2+3x+6\right)}{\left(x+1\right)\left[\sqrt[3]{\left(x^2+7\right)^2}+\sqrt[3]{x^2+7}\left(x+1\right)+\left(x+1\right)^2\right]}\)
\(=\dfrac{-\left(1+3+6\right)}{\left(1+1\right)\left(4+2.2+4\right)}=\dfrac{-5}{12}\)
Suy ra : \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}=\dfrac{-7}{8}+\dfrac{5}{12}=\dfrac{-11}{24}\)
1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)
2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0
3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)
\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)
4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)
a/ L'Hospital:
\(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)
b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)
b/ \(\lim\limits_{x\rightarrow-3}\dfrac{\left(x+2\right)\left(x+3\right)}{x\left(x+3\right)}=\lim\limits_{x\rightarrow-3}\dfrac{x+2}{x}=\dfrac{-3+2}{-3}=\dfrac{1}{3}\)
a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)
\(=\dfrac{2x-5}{7}\)
\(=\dfrac{2}{7}x-\dfrac{5}{7}\)
\(=-\infty\)
b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)
\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)
a: \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{2x+10-16}{3x-9}\cdot\dfrac{1}{\sqrt{2x+10}+4}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{2\left(x-3\right)}{3\left(x-3\right)\cdot\left(\sqrt{2x+10}+4\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{2}{3\left(\sqrt{2x+10}+4\right)}\)
\(=\dfrac{2}{3\cdot\sqrt{6+10}+3\cdot4}=\dfrac{2}{3\cdot4+3\cdot4}=\dfrac{2}{24}=\dfrac{1}{12}\)
b: \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)
\(=\lim\limits_{x\rightarrow7}\dfrac{4x+8-36}{\sqrt{4x+8}+6}\cdot\dfrac{1}{\left(x-2\right)\left(x-7\right)}\)
\(=\lim\limits_{x\rightarrow7}\dfrac{4x-28}{\left(\sqrt{4x+8}+6\right)\cdot\left(x-2\right)\left(x-7\right)}\)
\(=\lim\limits_{x\rightarrow7}\dfrac{4}{\left(\sqrt{4x+8}+6\right)\left(x-2\right)}\)
\(=\dfrac{4}{\left(\sqrt{4\cdot7+8}+6\right)\left(7-2\right)}\)
\(=\dfrac{4}{5\cdot12}=\dfrac{4}{60}=\dfrac{1}{15}\)
c: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{2x^2-10x+x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-5\right)\left(2x+1\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{x-3}{2x+1}=\dfrac{5-3}{2\cdot5+1}=\dfrac{2}{11}\)
a) Áp dụng giới hạn một bên thường dùng, ta có : \(\mathop {\lim }\limits_{x \to {4^ + }} \frac{1}{{x - 4}} = + \infty \)
b) \(\mathop {\lim }\limits_{x \to {2^ + }} \frac{x}{{2 - x}} = \mathop {\lim }\limits_{x \to {2^+ }} \frac{{ - x}}{{x - 2}} = \mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right).\mathop {\lim }\limits_{x \to {2^ + }} \frac{1}{{x - 2}}\)
Ta có: \(\mathop {\lim }\limits_{x \to {2^ + }} \left( { - x} \right) = - \mathop {\lim }\limits_{x \to {2^ + }} x = - 2;\mathop {\lim }\limits_{x \to {2^ +}} \frac{1}{{x - 2}} = +\infty \)
\( \Rightarrow \mathop {\lim }\limits_{x \to {2^ - }} \frac{x}{{2 - x}} = - \infty \)
