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\(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\Rightarrow f\left(2\right)=-1\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{f\left(x\right)+2x+1}+x}.\dfrac{\left(\sqrt{f\left(x\right)+2x+1}-x\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\dfrac{f\left(x\right)+1-x\left(x-2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\left(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}-\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{x-2}\right)\)
\(=\dfrac{1}{4\left(\sqrt{4}+2\right)}.\left(a-2\right)=\dfrac{a-2}{16}\)
Em kiểm tra lại đề, chỗ \(f\left(x\right)-32\) kia có vẻ sai, vì như thế thì biểu thức đã cho ko phải dạng vô định
Do \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-3}{x-2}=5\Rightarrow\) chọn \(f\left(x\right)=5\left(x-2\right)+3=5x-7\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-7+6}-\sqrt[3]{x+25}}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[]{5x-1}-3+3-\sqrt[3]{x+25}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{5\left(x-2\right)}{\sqrt[]{5x-1}+3}-\dfrac{x-2}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{5}{\sqrt[]{5x-1}+3}-\dfrac{1}{9+3\sqrt[3]{x+25}+\sqrt[3]{\left(x+25\right)^2}}\right)=\dfrac{5}{3+3}-\dfrac{1}{9+9+9}=\dfrac{43}{54}\)
a. Áp dụng công thức L'Hospital:
\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{1-x}}{\sqrt[3]{x+1}-\sqrt{1-x}}=\lim\limits_{x\to 0}\frac{\frac{1}{2}(x+1)^{\frac{-1}{2}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}{\frac{1}{3}(x+1)^{\frac{-2}{3}}+\frac{1}{2}(1-x)^{\frac{-1}{2}}}=\frac{1}{\frac{5}{6}}=\frac{6}{5}\)
b.
\(\lim\limits_{x\to 0}(\frac{1}{x}-\frac{1}{x^2})=\lim\limits_{x\to 0}\frac{x-1}{x^2}=-\infty\)
c. Áp dụng quy tắc L'Hospital:
\(\lim\limits_{x\to +\infty}\frac{x^4-x^3+11}{2x-7}=\lim\limits_{x\to +\infty}\frac{4x^3-3x^2}{2}=+\infty \)
d.
\(\lim\limits_{x\to 5}\frac{7}{(x-1)^2}.\frac{2x+1}{2x-3}=\frac{7}{(5-1)^2}.\frac{2.5+11}{2.5-3}=\frac{11}{16}\)
a/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\left(2x\right)^2.\left(4x\right)^3}{x^4}}{\dfrac{\left(3x\right)^2\left(5x^2\right)}{x^4}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{4^4.x}{45}=\pm\infty\)
b/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{2x^2}{x^3}+\dfrac{x}{x^3}}}{\dfrac{2x}{x}-\dfrac{2}{x}}=\dfrac{1}{2}\)
c/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\sqrt[3]{\left(x^3+2x^2\right)^2}}{x^2}+\dfrac{x\sqrt[3]{x^3+2x^2}}{x^2}+\dfrac{x^2}{x^2}}{\dfrac{3x^2}{x^2}-\dfrac{2x}{x^2}}=\dfrac{1+1+1}{3}=1\)
d/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(-3x\right)^3x^2}{x^5}}{-\dfrac{4x^5}{x^5}}=\dfrac{-27}{-4}=\dfrac{27}{4}\)
e/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(2x\right)^{20}.\left(3x\right)^{20}}{x^{50}}}{\dfrac{\left(2x\right)^{50}}{x^{50}}}=0\)
g/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{8x^3.\left(4x^5\right)^9}{x^{47}}}{\dfrac{11x^{47}}{x^{47}}}=+\infty\)
Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.
\(\lim\limits_{x\rightarrow+\infty}\dfrac{\left(5-3x\right)^9}{\left(2+x\right)^6}=\lim\limits_{x\rightarrow+\infty}\dfrac{x^6\left(\dfrac{5}{x}-3\right)^6\left(5-3x\right)^3}{x^6\left(\dfrac{2}{x}+1\right)^6}=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{5}{x}-3}{\dfrac{2}{x}+1}\right)^6\left(\dfrac{5}{x}-3\right)^3x^3\)
Do \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{5}{x}-3}{\dfrac{2}{x}+1}\right)^6\left(\dfrac{5}{x}-3\right)^3=\left(-\dfrac{3}{1}\right)^6\left(-3\right)^3=-3^9< 0\\\lim\limits_{x\rightarrow+\infty}x^3=+\infty\end{matrix}\right.\)
\(\Rightarrow\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{5}{x}-3}{\dfrac{2}{x}+1}\right)^6\left(\dfrac{5}{x}-3\right)^3x^3=-\infty\)
Em là tám lại ạ
Em là duy khôi ạ
Em là văn tam ạ
Em là mạnh Tuấn ạ
a: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}\)
\(=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}\)
b: \(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\left(\dfrac{x-3+x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2\)
d: \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x+x-\sqrt[3]{x^3-1}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\dfrac{x^3-x^3+1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{x^2+1}+x}+\dfrac{1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\dfrac{1}{x}}{\sqrt{1+\dfrac{1}{x^2}}+1}+\dfrac{\dfrac{1}{x^2}}{\sqrt[3]{\dfrac{1}{x^4}}+\sqrt[3]{1-\dfrac{1}{x^3}}+\sqrt[3]{\left(1-\dfrac{1}{x^3}\right)^2}}\right)\)
=0
c: \(\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right]\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}\)
e: \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^3-3x+2}{x^4-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{\left(x+2\right)\left(x-1\right)^2}{\left(x^2+2x+3\right)\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\dfrac{x+2}{x^2+2x+3}=\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow2^-}\dfrac{x^3+x^2-4x-4}{x^2-4x+4}=\lim\limits_{x\rightarrow2^-}\dfrac{\left(x-2\right)\left(x^2+3x+2\right)}{\left(x-2\right)^2}=\lim\limits_{x\rightarrow2^-}\dfrac{x^2+3x+2}{x-2}=-\infty\)
\(\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)^{20}}{\left(x^3-12x+16\right)^{10}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}\left(x-2\right)^{20}}{\left(x+4\right)^{10}\left(x-2\right)^{20}}=\lim\limits_{x\rightarrow2}\dfrac{\left(x+1\right)^{20}}{\left(x+4\right)^{10}}=\dfrac{3^{10}}{2^{10}}\)
\(\lim\limits_{x\rightarrow0^-}\dfrac{4x^2+5x}{x^2}=\lim\limits_{x\rightarrow0^-}\dfrac{4x+5}{x}=-\infty\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+2}-1}{\sqrt{x+5}-2}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(\sqrt{x+5}+2\right)}{\left(x+1\right)\left(\sqrt{x+2}+1\right)}=\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{x+5}+2}{\sqrt{x+2}+1}=2\)
ta có (f(x)-20)/(x-2)=10
===>f(x)=10x
thay f(x)=10x vào A và thay
x=2+0,000000001 ta được giới hạn của A= -331259694,9
cái chỗ F(x) =10x đó ,đâu có là sao vậy ạ , tại có thể 10 đó là g(2)=10