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\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}\) hữu hạn nên \(f\left(x\right)-16=0\) có nghiệm \(x=1\)
\(\Rightarrow f\left(1\right)=16\)
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}.\dfrac{1}{\sqrt{2f\left(x\right)+4}+6}=24.\dfrac{1}{\sqrt{2.16+4}+6}=2\)
Do \(\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5\) hữu hạn nên \(2f\left(x\right)+1=0\) phải có nghiệm \(x=-1\)
\(\Leftrightarrow2f\left(-1\right)=-1\Leftrightarrow f\left(-1\right)=-\dfrac{1}{2}\)
Đoạn dưới tự hiểu là \(\lim\limits_{x\rightarrow-1}\) (vì kí tự lim rất rắc rối)
\(I=\dfrac{\left[4f\left(x\right)+3\right]\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}-2\right]+2\left[4f\left(x\right)+3\right]-2}{x^2-1}\)
\(=\dfrac{\left[4f\left(x\right)+3\right]\left[4f^2\left(x\right)+2f\left(x\right)\right]}{\left(x+1\right)\left(x-1\right)\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}+2\right]}+\dfrac{4\left[2f\left(x\right)+1\right]}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{f\left(x\right).\left[4f\left(x\right)+3\right]}{x-1}+\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{4}{x-1}\)
\(=5.\dfrac{f\left(-1\right).\left[4f\left(-1\right)+3\right]}{-2}+5.\dfrac{4}{-2}=\dfrac{5.\left(-\dfrac{1}{2}\right)\left(-2+3\right)}{-2}+5.\dfrac{4}{-2}=...\)
Không phải dạng, nó chỉ là ứng dụng kiến thức cơ bản về giới hạn của hàm thôi
\(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}\) hữu hạn \(\Rightarrow f\left(x\right)+1=0\) có nghiệm \(x=2\Rightarrow f\left(2\right)=-1\)
\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{f\left(x\right)+2x+1}-x}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{1}{\sqrt{f\left(x\right)+2x+1}+x}.\dfrac{\left(\sqrt{f\left(x\right)+2x+1}-x\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\dfrac{f\left(x\right)+1-x\left(x-2\right)}{x-2}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x+2\right)\left(\sqrt{f\left(x\right)+2x+1}+x\right)}.\left(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)+1}{x-2}-\lim\limits_{x\rightarrow2}\dfrac{x\left(x-2\right)}{x-2}\right)\)
\(=\dfrac{1}{4\left(\sqrt{4}+2\right)}.\left(a-2\right)=\dfrac{a-2}{16}\)
Do \(\lim\limits_{x\rightarrow3}\dfrac{f\left(x\right)-2}{x-3}\) hữu hạn \(\Rightarrow f\left(x\right)-2=0\) có nghiệm \(x=3\)
Hay \(f\left(3\right)-2=0\Rightarrow f\left(3\right)=2\)
\(\Rightarrow I=\lim\limits_{x\rightarrow3}\left(\dfrac{f\left(x\right)-2}{x-3}\right).\dfrac{1}{\sqrt{5f\left(x\right)+6}+1}=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.f\left(3\right)+6}+1}\)
\(=\dfrac{1}{4}.\dfrac{1}{\sqrt{5.2+6}+1}=\dfrac{1}{20}\)
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
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:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
thì f(x) thỏa mãn được tất cả các điều kiện đã nêu