Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giới hạn này tiến đến đâu vậy bạn? 2 trường hợp khác nhau đúng ko?
\(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}=\lim\limits_{x\rightarrow+\infty}\frac{x\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=1\)
\(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+3x+5}}{\sqrt[3]{x^3+7x^2+8}}=\lim\limits_{x\rightarrow-\infty}\frac{\left|x\right|\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=\lim\limits_{x\rightarrow-\infty}\frac{-x\sqrt{1+\frac{3}{x}+\frac{5}{x^2}}}{x\sqrt[3]{1+\frac{7}{x}+\frac{8}{x^3}}}=-1\)
Hai trường hợp sẽ cho ra 2 kết quả khác nhau bạn
Lời giải:
\(\lim\limits_{x\to \pm\infty}\sqrt{x^2-3x+4}=\lim\limits_{x\to \pm\infty}\sqrt{x^2}.\lim\limits_{x\to \pm \infty}\sqrt{1-\frac{3}{x}+\frac{4}{x^2}}=\lim\limits_{x\to \pm\infty}|x|.1=+\infty \)
--------------
\(\lim\limits_{x\to +\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to +\infty}x^2.\lim\limits_{x\to +\infty}(\sqrt{1+\frac{5}{x^2}}+1)=2(+\infty )=+\infty \)
\(\lim\limits_{x\to -\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to -\infty}\frac{5x}{\sqrt{x^2+5}-x}=\lim\limits_{x\to -\infty}\frac{-5}{\sqrt{1+\frac{5}{x^2}}+1}=\frac{-5}{2}\)
----------------
\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
Lời giải:
\(\lim\limits_{x\to \pm\infty}\sqrt{x^2-3x+4}=\lim\limits_{x\to \pm\infty}\sqrt{x^2}.\lim\limits_{x\to \pm \infty}\sqrt{1-\frac{3}{x}+\frac{4}{x^2}}=\lim\limits_{x\to \pm\infty}|x|.1=+\infty \)
--------------
\(\lim\limits_{x\to +\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to +\infty}x^2.\lim\limits_{x\to +\infty}(\sqrt{1+\frac{5}{x^2}}+1)=2(+\infty )=+\infty \)
\(\lim\limits_{x\to -\infty}x(\sqrt{x^2+5}+x)=\lim\limits_{x\to -\infty}\frac{5x}{\sqrt{x^2+5}-x}=\lim\limits_{x\to -\infty}\frac{-5}{\sqrt{1+\frac{5}{x^2}}+1}=\frac{-5}{2}\)
----------------
\(\lim\limits_{x\to 2019}\frac{\sqrt{x+285}-48}{\sqrt{x-2018}-\sqrt{2020-x}}=\lim\limits_{x\to -\infty}(\sqrt{x+285}-48).\lim\limits_{x\to -\infty}\frac{1}{\sqrt{x-2018}-\sqrt{2020-x}}\)
\(=\lim\limits_{x\to 2019}\frac{x-2019}{\sqrt{x+285}+48}.\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(x-2019)}=\lim\limits_{x\to 2019}\frac{\sqrt{x-2018}+\sqrt{2020-x}}{2(\sqrt{x+285}+48)}=\frac{1}{96}\)
Đây là giới hạn dạng vô định \(\frac{0}{0}\)
\(=\lim\limits_{x\rightarrow3^+}\frac{\sqrt{x-3}.\sqrt{\left(x+1\right)\left(x-3\right)}}{\sqrt{\left(x+3\right)\left(x-3\right)}}=\lim\limits_{x\rightarrow3^+}\frac{\sqrt{\left(x+1\right)\left(x-3\right)}}{\sqrt{x+3}}=\frac{0}{\sqrt{6}}=0\)
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/ \(lim\frac{\sqrt{-x}\left(2\sqrt{-x}+1\right)}{\sqrt{-x}\left(5\sqrt{-x}-1\right)}=lim\frac{2\sqrt{-x}+1}{5\sqrt{-x}-1}=\frac{1}{-1}=-1\)
1/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}.\sqrt[4]{1+8x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}.\sqrt[4]{1+8x}-\sqrt[3]{1+6x}}{x}+\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{1+6x}-1}{x}\)
Liên hợp dài quá ko muốn gõ tiếp, bạn tự đặt nhân tử chung rồi liên hợp nhé, kết quả ra 5
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{1+7x}-2-\left(x^3-3x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{7\left(x-1\right)}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)^2\left(x+2\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{7}{\sqrt[3]{\left(1+7x\right)^2}+2\sqrt[3]{1+7x}+4}-\left(x-1\right)\left(x+2\right)=\dfrac{7}{12}\)
3/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x^3-x^2+1}{2x^2+3x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x-1+\dfrac{1}{x^2}}{2+\dfrac{3}{x}-\dfrac{1}{x^2}}=-\infty\)
4/ \(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}}{\sqrt{4x+1}}=\lim\limits_{x\rightarrow+\infty}\dfrac{1+\dfrac{1}{\sqrt[6]{x}}+\dfrac{1}{\sqrt[4]{x}}}{\sqrt{4+\dfrac{1}{x}}}=\dfrac{1}{\sqrt{4}}=\dfrac{1}{2}\)
5/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+\sqrt{x^2+2}}{\sqrt[3]{8x^3+x^2+1}}=\lim\limits_{x\rightarrow-\infty}\dfrac{1-\sqrt{1+\dfrac{2}{x^2}}}{\sqrt[3]{8+\dfrac{1}{x}+\dfrac{1}{x^3}}}=\dfrac{1-1}{\sqrt[3]{8}}=0\)
6/ \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{4x^2+3x-7}}{\sqrt[3]{27x^3+5x^2+x-4}}=\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{4+\dfrac{3}{x}-\dfrac{7}{x^2}}}{\sqrt[3]{27+\dfrac{5}{x}+\dfrac{1}{x^2}-\dfrac{4}{x^3}}}=\dfrac{-\sqrt{4}}{\sqrt[3]{27}}=\dfrac{-2}{3}\)
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
Câu dưới là 1 giới hạn hoàn toàn bình thường (không phải dạng vô định), bạn cứ thay số vào là được thôi
\(\lim\limits_{x\rightarrow0}\left(1-x\right)tan\frac{\pi x}{2}=\left(1-0\right).tan0=1\)
Đây không phải giới hạn dạng vô định mà chỉ là giới hạn bình thường
\(=\frac{\sqrt[3]{19}-2\sqrt{2}}{0}=-\infty\)
ok ~