\(=\lim\limits_{x\rightarrow\dfrac{1}{2}^-}\dfrac{4.\left(\dfrac{1}{2}\right)+1}{0}=+\infty\)
\(=\lim\limits_{x\rightarrow\dfrac{1}{2}^-}\dfrac{4.\left(\dfrac{1}{2}\right)+1}{0}=+\infty\)
\(\lim\limits_{\rightarrow-\infty}\left(2x-1\right)\sqrt{\dfrac{x-1}{4x^2+x-5}}\)
Tìm giới hạn:
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)
Tính đạo hàm:
1. \(\lim\limits_{x\rightarrow3^+}\sqrt{x^2-9}.\dfrac{2x+1}{x-3}\)
2. \(\lim\limits_{x\rightarrow2^-}\left(x^3-8\right)\sqrt{\dfrac{x}{2-x^2}}\)
Tính giới hạn
\(\overset{Lim}{x\rightarrow\dfrac{\pi}{6}}\dfrac{2\sin x-1}{6x-\pi}\)
\(\lim\limits_{\rightarrow+\infty}\dfrac{2016x^2-2014x+2012}{2017x^2+2015x+2013}\)
giúp em với ạ làm mãi không ra ạ
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
đạo hàm
a) \(y=\sqrt{\dfrac{2x-1}{x+1}}\)
b) \(y=4x+\dfrac{3}{2}x^2\)
c) \(y=\dfrac{x^3}{3}-4x^2+7x+1\)
Tính đạo hàm:
1) \(y = \sin^2 \sqrt {4x+3}\)
2) \(y = \dfrac{3}{4}x^4 - \dfrac{34}{\sqrt{x}} + \pi\)
3) \(y = \sqrt{\dfrac{\sin4x}{\cos(x^2+2)}}\)
4) \(y = \dfrac{1}{\sqrt{\sin^2(6-x)+4x}}\)
5) \(y = x.\sin^2\left(\dfrac{2x-1}{4-x}\right)\)
6) \(y = \dfrac{4}{3}x^3 + \dfrac{3}{2\sqrt{x}} + \sqrt{2x}\)
7) \(y = \sqrt{\cot^3(x^2-1)} + \left(\dfrac{\sin2x}{\cos3x}\right)^4\)
8) \(y = \dfrac{\tan3x}{\cot^23x} - (\sin2x + \cos3x)^5\)
9) \(y = \cot^65x - \cos^43x + \sin3x\)
Tính đạo hàm:
a) y= \(\dfrac{x^3+2\sqrt{x-1}}{x-1}\)
b) y= \(\dfrac{4x^3+2x-3}{\sqrt{x^2+2}}\)
c) y= \(|x^3+x+1|\)
d) y= \(\sqrt{7-6x^4+x^3}\)
e) y= \(\dfrac{x^5+1}{2-\sqrt{x^2+3}}\)