Dùng định nghĩa tìm các giới hạn :
a) \(\lim\limits_{x\rightarrow5}\dfrac{x+3}{3-x}\)\(\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x^3+1}{x^2+1}\)
Tìm các giới hạn sau:
\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)
\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)
\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)
Dùng định nghĩa, tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow4}\dfrac{x+1}{3x-2}\)
b) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2-5x^2}{x^2+3}\)
a) Hàm số f(x) = xác định trên R\{} và ta có x = 4 ∈ (;+∞).
Giả sử (xn) là dãy số bất kì và xn ∈ (;+∞); xn ≠ 4 và xn → 4 khi n → +∞.
Ta có lim f(xn) = lim = = .
Vậy = .
b) Hàm số f(x) = xác định trên R.
Giả sử (xn) là dãy số bất kì và xn → +∞ khi n → +∞.
Ta có lim f(xn) = lim = lim = -5.
Vậy = -5.
Tính các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-3}\dfrac{x+3}{x^2+2x-3}\)
b) \(\lim\limits_{x\rightarrow0}\dfrac{\left(1+x\right)^3-1}{x}\)
c) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-1}{x^2-1}\)
d) \(\lim\limits_{x\rightarrow5}\dfrac{x-5}{\sqrt{x}-\sqrt{5}}\)
e) \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-5}{\sqrt{x}+\sqrt{5}}\)
f) \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x^2+5}-3}{x+2}\)
g) \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{x}-1}{\sqrt{x+3}-2}\)
h) \(\lim\limits_{x\rightarrow+\infty}\dfrac{1-2x+3x^3}{x^3-9}\)
i) \(\lim\limits_{x\rightarrow0}\dfrac{1}{x^2}\left(\dfrac{1}{x^2+1}-1\right)\)
j) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(x^2-1\right)\left(1-2x\right)^5}{x^7+x+3}\)
Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow0^-}\dfrac{2\left|x\right|+x}{x^2-x}\)
b) \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt[3]{1+x^4+x^6}}{\sqrt{1+x^3+x^4}}\)
a: \(\lim\limits_{x->0^-^-}\dfrac{-2x+x}{x\left(x-1\right)}=lim_{x->0^-}\left(\dfrac{-x}{x\left(x-1\right)}\right)\)
\(=lim_{x->0^-}\left(\dfrac{-1}{x-1}\right)=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)
b: \(=lim_{x->-\infty}\left(\dfrac{x^2-x-x^2+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-x+1}{\sqrt{x^2-x}+\sqrt{x^2-1}}\right)\)
\(=lim_{x->-\infty}\left(\dfrac{-1+\dfrac{1}{x}}{-\sqrt{1-\dfrac{1}{x^2}}-\sqrt{1-\dfrac{1}{x^2}}}\right)=\dfrac{-1}{-2}=\dfrac{1}{2}\)
BÀI 3. Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^3-5x^2+1}{7x^2-x+4}\)
b) \(\lim\limits_{x\rightarrow+\infty}x\sqrt{\dfrac{x^2+2x+3}{3x^4+4x^2-5}}\)
a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)
\(=\dfrac{2x-5}{7}\)
\(=\dfrac{2}{7}x-\dfrac{5}{7}\)
\(=-\infty\)
b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)
\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)
tính giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\dfrac{5x^2+x^3+5}{4x^3+1}\)
b) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
c) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2-x+1}{x^3+x-2x^2}\)
`a)lim_{x->+oo}[5x^2+x^3+5]/[4x^3+1]` `ĐK: 4x^3+1 ne 0`
`=lim_{x->+oo}[5/x+1+5/[x^3]]/[4+1/[x^3]]`
`=1/4`
`b)lim_{x->-oo}[2x^2-x+1]/[x^3+x-2x^2]` `ĐK: x ne 0;x ne 1`
`=lim_{x->-oo}[2/x-1/[x^2]+1/[x^3]]/[1+1/[x^2]-2/x]`
`=0`
Câu `c` giống `b`.
Tìm các giới hạn sau :
a) \(\lim\limits_{x\rightarrow2}\dfrac{x+3}{x^2+x+4}\)
b) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2+5x+6}{x^2+3x}\)
c) \(\lim\limits_{x\rightarrow4^-}\dfrac{2x-5}{x-4}\)
d) \(\lim\limits_{x\rightarrow+\infty}\left(-x^3+x^2-2x+1\right)\)
e) \(\lim\limits_{x\rightarrow-\infty}\dfrac{x+3}{3x-1}\)
f) \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-2x+4}-x}{3x-1}\)
Tính các giới hạn sau :
a) \(\lim\limits_{x\rightarrow-3}\dfrac{x^2-1}{x+1}\)
b) \(\lim\limits_{x\rightarrow-2}\dfrac{4-x^2}{x+2}\)
c) \(\lim\limits_{x\rightarrow6}\dfrac{\sqrt{x+3}-3}{x-6}\)
d) \(\lim\limits_{x\rightarrow+\infty}\dfrac{2x-6}{4-x}\)
e) \(\lim\limits_{x\rightarrow+\infty}\dfrac{17}{x^2+1}\)
f) \(\lim\limits_{x\rightarrow+\infty}\dfrac{-2x^2+x-1}{3+x}\)
a) = = -4.
b) = = (2-x) = 4.
c) =
= = = .
d) = = -2.
e) = 0 vì (x2 + 1) = x2( 1 + ) = +∞.
f) = = -∞, vì > 0 với ∀x>0.
Tìm giới hạn:
a, \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-2}{3-\sqrt{x^2+7}}\)
b, \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x}-\sqrt{4x^2+1}}{2x+3}\)
a: \(\lim\limits_{x\rightarrow+\infty}\dfrac{x-2}{3-\sqrt{x^2+7}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{1-\dfrac{2}{x}}{\dfrac{3}{x}-\sqrt{1+\dfrac{7}{x^2}}}\)
\(=\dfrac{1}{0-\sqrt{1+0}}=\dfrac{1}{-1}=-1\)
b: \(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2-x}-\sqrt{4x^2+1}}{2x+3}\)
\(=\dfrac{\sqrt{x^2\left(1-\dfrac{1}{x}\right)}-\sqrt{x^2\left(4+\dfrac{1}{x^2}\right)}}{2x+3}\)
\(=\dfrac{-x\cdot\sqrt{1-\dfrac{1}{x}}+x\cdot\sqrt{4+\dfrac{1}{x^2}}}{x\left(2+\dfrac{3}{x}\right)}\)
\(=\dfrac{-\sqrt{1-\dfrac{1}{x}}+\sqrt{4+\dfrac{1}{x^2}}}{2+\dfrac{3}{x}}=\dfrac{-1+2}{2}=\dfrac{1}{2}\)