Tìm các giới hạn sau:
a) \(lim_{x\rightarrow0}\dfrac{tan3x}{sin5x}\)
b) \(lim_{x\rightarrow0}\dfrac{cos2x-1}{sin^23x}\)
c) \(lim_{x\rightarrow1}\dfrac{x^2-4x+3}{sin\left(x-1\right)}\)
Giới hạn nào sau đây tồn tại:
A, \(lim_{x\rightarrow+\infty}sin2x\) B, \(lim_{x\rightarrow+\infty}cos3x\) C, \(lim_{x\rightarrow0}sin\frac{1}{2x}\) D, \(lim_{x\rightarrow1}sin\frac{1}{2x}\)
Đáp án D đúng
\(lim_{x\rightarrow0}\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)\)
\(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{x}-\dfrac{1}{x^2}\right)\)
\(=\lim\limits_{x\rightarrow0}\dfrac{x-1}{x^2}\)
\(=-\infty\) vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow0}x-1=0-1=-1< 0\\\lim\limits_{x\rightarrow0}x^2=0^2=0\end{matrix}\right.\)
Tìm các giới hạn sau:
a) \(\lim\limits_{h\rightarrow0}\dfrac{2\left(x+h\right)^3-2x^3}{h}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{\left(x+x^2+...+x^{2021}\right)-2021}{x-1}\)
a/ \(=\lim\limits_{h\rightarrow0}\dfrac{2x^3+6x^2h+6xh^2+2h^3-2x^3}{h}\)
\(=\lim\limits_{h\rightarrow0}\dfrac{6xh^2+6x^2h+2h^3}{h}=\lim\limits_{h\rightarrow0}\left(6xh+6x^2+2h^2\right)=6x^2\)
b/ Xet day :\(S=x+x^2+....+x^{2021}\)
Day co \(\left\{{}\begin{matrix}u_1=x\\q=x\end{matrix}\right.\Rightarrow S=u_1.\dfrac{q^{2021}-1}{q-1}=x.\dfrac{x^{2021}-1}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}-x}{x-1}-2021}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-x-2021x+2021}{\left(x-1\right)^2}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^{2022}}{x^2}-\dfrac{x}{x^2}-\dfrac{2021x}{x^2}+\dfrac{2021}{x^2}}{\dfrac{x^2}{x^2}-\dfrac{2x}{x^2}+\dfrac{1}{x^2}}=\lim\limits_{x\rightarrow1}\dfrac{x^{2020}}{1}=1\)
Lam lai cau b, hinh nhu bi nham sang dang \(\dfrac{\infty}{\infty}\) roi
Xet day: \(S=x+x^2+...+x^{2021}\)
\(\Rightarrow S=x.\dfrac{x^{2021}-1}{x-1}=\dfrac{x^{2022}-x}{x-1}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\dfrac{x^{2022}-2022x+2021}{\left(x-1\right)^2}\)
L'Hospital: \(\Rightarrow...=\lim\limits_{x\rightarrow1}\dfrac{2022x^{2021}-2022}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\dfrac{2022.2021.x^{2020}}{2}=2043231\)
Is that true :v?
Cau a co the xai L'Hospital cung ra:
L'Hospital:
\(...=\lim\limits_{h\rightarrow0}\dfrac{6xh^2+6x^2h+2h^3}{h}=\lim\limits_{h\rightarrow0}\dfrac{6h^2+12xh+6x^2+12xh+6h^2}{1}=6x^2\)
\(lim_{x\rightarrow0}\dfrac{\sqrt{x^3+1}-1}{x^2+x}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^3+1}-1}{x^2+x}\\ =\lim\limits_{x\rightarrow0}\dfrac{x\sqrt{x+\dfrac{1}{x^2}}-1}{x\left(x+1\right)}\\ =\lim\limits_{x\rightarrow0}\dfrac{x\left(\sqrt{x+\dfrac{1}{x^2}}-\dfrac{1}{x}\right)}{x\left(x+1\right)}\\ =\dfrac{\sqrt{x+\dfrac{1}{x^2}}-\dfrac{1}{x}}{x+1}\\ =\dfrac{\sqrt{x}}{x+1}=\dfrac{0}{0+1}=0\)
Cho \(f\left(x\right)\) xác định trên khoảng nào đó chứa điểm 0 và \(\left|f\left(x\right)\right|\le\left|x\right|\) . Khi đó ta có:
A, \(lim_{x\rightarrow0}f\left(x\right)=0\) B, \(lim_{x\rightarrow0}f\left(x\right)=1\) C, \(lim_{x\rightarrow0}f\left(x\right)=-1\) D, Hàm số không có giới hạn tại không.
Đáp án A
Đó là nguyên lý của giới hạn kẹp
\(\left|f\left(x\right)\right|\le\left|x\right|\Rightarrow\lim\limits_{x\rightarrow0}f\left(x\right)=\lim\limits_{x\rightarrow0}x=0\)
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: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow a}\dfrac{\sin x-\sin a}{x-a}\)
b) \(\lim\limits_{x\rightarrow1}\left(1-x\right)\tan\dfrac{\pi x}{2}\)
c) \(\lim\limits_{x\rightarrow\dfrac{\pi}{3}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}\)
d) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
\(lim_{x->a}\left[\dfrac{1}{\left(x-a\right)^2}\left(x^2-8x+10+\dfrac{81}{x+2\sqrt{x-1}}-2\sqrt{x-1}\right)\right]=\dfrac{21}{16}\)
\(lim_{x->b}\left[\dfrac{4}{\left(x-b\right)^2}\left(x^2-x+2-2\sqrt{x}\right)\right]=c\)
với a,b,c là các số thực. Tìm a,b,c
Tính giới hạn L=\(lim_{x\rightarrow0}\frac{\left(1+x\right)^n-1}{x}\).Với n là số nguyên dương
\(lim_{x\rightarrow0+}\frac{\left(1+x\right)^n-1}{x}\)
\(=lim_{x\rightarrow0+}\frac{\left(1+x\right)^n-1^n}{x}\)
\(=lim_{x\rightarrow0+}\frac{\left(1+x-1\right)\left[\left(1+x\right)^{n-1}+\left(1+x\right)^{n-2}+...+\left(1+x\right)^0\right]}{x}\)
\(=lim_{x\rightarrow0}\left[\left(1+x\right)^{n-1}+\left(1+x\right)^{n-2}+...\left(1+x\right)^0\right]\)
\(=1^{n-1}+1^{n-2}+...+1^0\)
Số số hạng
\(\left(n-1-0\right):1+1=n\)
Do mọi số hạng đều bằng 1 nên tổng là
\(1\cdot n=n\)