tìm giới hạn
\(\lim\limits_{x\rightarrow1}\frac{1+sin\pi x}{x+1}\)
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) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
Tìm các giới hạn :
a) \(\lim\limits_{x\rightarrow1}\dfrac{x^2-5x+6}{x-2}\)
b) \(\lim\limits_{x\rightarrow\dfrac{\pi}{8}}\dfrac{\sin2x-\cos2x}{8x-\pi}\)
a/ \(\lim\limits_{x\to 1} f(x)=\frac{x^{2}-5x + 6}{x-2} \)
\(<=>\lim\limits_{x\to 1} f(x)=\dfrac{(x-3)(x-2)}{x-2} \)
<=>\(\lim\limits_{x\to 1} f(x)=x-3 \)
\(<=>\lim\limits_{x\to 1} f(x)=-2\)
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow\frac{\pi}{2}}\frac{sin\left(x-\frac{\pi}{4}\right)}{x}\)
b, \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{3x^2-4}-\sqrt{3x-2}}{x+1}\)
c,\(\lim\limits_{x\rightarrow0}x^2sin\frac{1}{2}\)
Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:
\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)
\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)
\(c=0.sin\frac{1}{2}=0\)
Tìm giới hạn:
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[4]{x}-1}{x^3+x-2}\)
Tìm giới hạn bên phải: \(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x}+\sqrt{x-1}-1}{\sqrt{x^2-1}}\)
\(\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x}+\sqrt{x-1}-1}{\sqrt{x^2-1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\dfrac{\left(x-1\right)}{\sqrt{x}+1}+\left(\sqrt{x-1}\right)}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\sqrt{x-1}\right)\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x-1}\cdot\sqrt{x+1}}\)
\(=\lim\limits_{x\rightarrow1^+}\dfrac{\left(\dfrac{\sqrt{x-1}}{\sqrt{x}+1}+1\right)}{\sqrt{x+1}}=\dfrac{\dfrac{\sqrt{1-1}}{\sqrt{1}+1}+1}{\sqrt{1+1}}\)
\(=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
Bài 1. Tìm các giới hạn sau:
a) \(\lim\limits\dfrac{-2n+1}{n}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}\)
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
b; \(\lim\limits_{x\rightarrow1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c; \(\lim\limits_{x\rightarrow1}\frac{x^3-2x-1}{x^5-2x-1}\)
d; \(\lim\limits_{x\rightarrow-1}\frac{\left(x+2\right)^2-1}{x^2-1}\)
b.
\(\lim\limits_{x\to 1+}\frac{x^3-3x^2+2}{x^4-4x+3}=\lim\limits_{x\to 1+}\frac{(x-1)(x^2-2x-2)}{(x-1)^2(x^2+2x+3)}=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{(x-1)(x^2+2x+3)}\)
\(=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{x^2+2x+3}.\lim\limits_{x\to 1+}\frac{1}{x-1}=\frac{-1}{2}.(+\infty)=-\infty \)
Tương tự \(\lim\limits_{x\to 1-}\frac{x^3-3x^2+2}{x^4-4x+3}=+\infty \)
Do đó không tồn tại \(\lim\limits_{x\to 1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c.
\(\lim\limits_{x\to 1}\frac{x^3-2x-1}{x^5-2x-1}=\frac{1^3-2.1-1}{1^5-2.1-1}=1\)
d.
\(\lim\limits_{x\to -1}\frac{(x+2)^2-1}{x^2-1}=\lim\limits_{x\to -1}\frac{(x+2-1)(x+2+1)}{(x-1)(x+1)}=\lim\limits_{x\to -1}\frac{x+3}{x-1}=-1\)
a.
\(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=\lim_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{(x-1)^3(3x+1)}=\lim\limits _{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x+1}.\lim\limits_{x\to 1+}\frac{1}{(x-1)^3}\)
\(=\frac{1}{4}.(+\infty)=+\infty \)
Hoàn toàn tương tự:
\(\lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}=-\infty \)
Do đó: \(\lim\limits_{x\to 1+}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\neq \lim\limits_{x\to 1-}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\) nên không tồn tại \(\lim\limits_{x\to 1}\frac{2x^4-5x^3+3x^2+1}{3x^4-8x^3+6x^2-1}\)
b.
\(\lim\limits_{x\to 1+}\frac{x^3-3x^2+2}{x^4-4x+3}=\lim\limits_{x\to 1+}\frac{(x-1)(x^2-2x-2)}{(x-1)^2(x^2+2x+3)}=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{(x-1)(x^2+2x+3)}\)
\(=\lim\limits_{x\to 1+}\frac{x^2-2x-2}{x^2+2x+3}.\lim\limits_{x\to 1+}\frac{1}{x-1}=\frac{-1}{2}.(+\infty)=-\infty \)
Tương tự \(\lim\limits_{x\to 1-}\frac{x^3-3x^2+2}{x^4-4x+3}=+\infty \)
Do đó không tồn tại \(\lim\limits_{x\to 1}\frac{x^3-3x^2+2}{x^4-4x+3}\)
c.
\(\lim\limits_{x\to 1}\frac{x^3-2x-1}{x^5-2x-1}=\frac{1^3-2.1-1}{1^5-2.1-1}=1\)
d.
\(\lim\limits_{x\to -1}\frac{(x+2)^2-1}{x^2-1}=\lim\limits_{x\to -1}\frac{(x+2-1)(x+2+1)}{(x-1)(x+1)}=\lim\limits_{x\to -1}\frac{x+3}{x-1}=-1\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow1^+}\dfrac{x^3+x+1}{x-1}\)
b) \(\lim\limits_{x\rightarrow-1^+}\dfrac{3x+2}{x+1}\)
c) \(\lim\limits_{x\rightarrow2^-}\dfrac{x-15}{x-2}\)
Lời giải:
a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)
\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$
\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)
b.
\(\lim\limits_{x\to -1+}(3x+2)=-1<0\)
\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$
\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)
c.
\(\lim\limits_{x\to 2-}(x-15)=-17<0\)
\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$
\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)
Tìm các giới hạn sau:
a) \(\lim\limits_{x\rightarrow2}\dfrac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}\)
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:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)