Cho \(\xrightarrow[x->1]{lim}\dfrac{f\left(x\right)-10}{x-1}=5.\)
Tính \(\xrightarrow[x->1]{lim}\dfrac{f\left(x\right)-10}{\left(\sqrt{x}-1\right)\left(\sqrt{4f\left(x\right)+9}+3\right)}\)
Bài 1:Cho \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-10}{x-1}=5\) ,\(g\left(x\right)=\sqrt{f\left(x\right)+6}-2\sqrt[3]{f\left(x\right)-2}\)
Tính \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt{x}-1\right)g\left(x\right)}\)
Bài 2: Cho \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2ax^2+30}-bx-5}{x^3-3x+2}=c\left(a;b;c\in R\right)\)
Tính giá trị \(P=a^2+b^2+36c\)
Bài 3: Cho a;b là các số nguyên dương. Biết \(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2+ax}+\sqrt[3]{8x^3+2bx^2+3}\right)=\dfrac{7}{3}\)
Tinh P= a+2b
Bài 4:Cho a,b,c thuộc R với a>0 thỏa mãn
\(c^2+a=2\) và \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{ax^2+bx}-cx\right)=-3\)
Tính P= a+b+5c
Bài 5:
Mấy câu này khó nên mong các bạn giúp mình với. Mai mình phải kiểm tra rồi
Mấy câu này bạn cần giải theo kiểu trắc nghiệm hay tự luận nhỉ?
Làm tự luận thì hơi tốn thời gian đấy (đi thi sẽ không bao giờ đủ thời gian đâu)
Câu 1:
Kiểm tra lại đề, \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}\) hay một trong 2 giới hạn sau: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{x}-1}{g\left(x\right)}\) hoặc \(\lim\limits_{x\rightarrow1}\dfrac{g\left(x\right)}{\sqrt[]{x}-1}\)
Vì đúng như đề của bạn thì \(\lim\limits_{x\rightarrow1}\dfrac{1}{\left(\sqrt[]{x}-1\right)g\left(x\right)}=\dfrac{1}{0}=\infty\), cả \(g\left(x\right)\) lẫn \(\sqrt{x}-1\) đều tiến tới 0 khi x dần tới 1
Cho f(x) thỏa mãn : \(_{\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5}\)
Tính I= \(\lim\limits_{x\rightarrow-1}\dfrac{\left(4f\left(x\right)+3\right)\left(\sqrt{4f\left(x\right)^2+2f\left(x\right)+4}\right)-2}{x^2-1}\)
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)
\(I=\dfrac{\left[4f\left(x\right)+3\right]\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}-2\right]+2\left[4f\left(x\right)+3\right]-2}{x^2-1}\)
\(=\dfrac{\left[4f\left(x\right)+3\right]\left[4f^2\left(x\right)+2f\left(x\right)\right]}{\left(x+1\right)\left(x-1\right)\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}+2\right]}+\dfrac{4\left[2f\left(x\right)+1\right]}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{f\left(x\right).\left[4f\left(x\right)+3\right]}{x-1}+\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{4}{x-1}\)
\(=5.\dfrac{f\left(-1\right).\left[4f\left(-1\right)+3\right]}{-2}+5.\dfrac{4}{-2}=\dfrac{5.\left(-\dfrac{1}{2}\right)\left(-2+3\right)}{-2}+5.\dfrac{4}{-2}=...\)
Tìm giới hạn
1) \(\xrightarrow[x->3]{lim}\dfrac{x^2-5x+6}{\sqrt{2x+3}-\sqrt{4x-3}}\)
2) \(\xrightarrow[x->1]{lim}\dfrac{\sqrt{x^2+2}-\sqrt{4x-1}}{x-1}\)
3) \(\xrightarrow[x->-1]{lim}\dfrac{x-2}{x\left|x+1\right|}\)
4) \(\xrightarrow[x->a]{lim}\dfrac{x^n-a^n}{x-a}\)
5) \(\xrightarrow[x->1]{lim}(\dfrac{n}{1-x^n}-\dfrac{1}{1-x})\)
6) \(\xrightarrow[x->1]{lim}\dfrac{x^n-nx+n-1}{\left(x-1\right)^2}\)
Cho \(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}=5\). Tính \(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3f\left(x\right)+10}+\sqrt{f^3\left(x\right)+1}-7}{x^2-25}\)
Chọn F(x)=5x-23
\(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5x-23-2}{x-5}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{5x-25}{x-5}=\lim\limits_{x\rightarrow5}\dfrac{5\left(x-5\right)}{x-5}=5\)
=>f(x)=5x-23 thỏa mãn yêu cầu đề bài
\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\cdot f\left(x\right)+10}+\sqrt{f^3\left(x\right)+1}-7}{x^2-25}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3\left(5x-23\right)+10}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}+\sqrt{\left(5x-23\right)^3+1}-7}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\sqrt{15x-59}-4+\sqrt{\left(5x-23\right)^3+1}-3}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15x-59-16}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3+1-9}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23\right)^3-8}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15\left(x-5\right)}{\sqrt{15x-59}+4}+\dfrac{\left(5x-23-2\right)\left[\left(5x-23\right)^2+2\left(5x-23\right)+4\right]}{\sqrt{\left(5x-23\right)^3+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{15}{\sqrt{15x-59}+4}+\dfrac{5\cdot\left(25x^2-230x+529+10x-46+4\right)}{\sqrt{\left(5x-23\right)^3+1}+3}}{x+5}\)
\(=\dfrac{\dfrac{15}{\sqrt{15\cdot5-59}+4}+\dfrac{5\left(25\cdot5^2-220\cdot5+487\right)}{\sqrt{\left(5\cdot5-23\right)^3+1}+3}}{5+5}\)
\(=\dfrac{\dfrac{15}{8}+\dfrac{5\cdot12}{6}}{10}=\dfrac{19}{16}\)
Do \(\lim\limits_{x\rightarrow5}\dfrac{f\left(x\right)-2}{x-5}\) hữu hạn nên \(f\left(x\right)-2=0\) có nghiệm \(x=5\)
\(\Rightarrow f\left(5\right)=2\)
\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{3f\left(x\right)+10}-4+\sqrt{f^3\left(x\right)+1}-3}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{3\left[f\left(x\right)-2\right]}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{\left[f\left(x\right)-2\right]\left[f^2\left(x\right)+2f\left(x\right)+4\right]}{\sqrt{f^3\left(x\right)+1}+3}}{\left(x-5\right)\left(x+5\right)}\)
\(=\lim\limits_{x\rightarrow5}\dfrac{\dfrac{f\left(x\right)-2}{x-5}.\dfrac{3}{\sqrt{3f\left(x\right)+10}+4}+\dfrac{f\left(x\right)-2}{x-5}.\dfrac{f^2\left(x\right)+2f\left(x\right)+4}{\sqrt{f^3\left(x\right)+1}+3}}{x+5}\)
\(=\dfrac{5.\dfrac{3}{\sqrt{3.2+10}+4}+5.\dfrac{2^2+2.2+4}{\sqrt{2^3+1}+3}}{5+5}=\)
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: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)
Cho \(\lim\limits_{x\rightarrow2}\dfrac{f\left(x\right)-32}{x-2}=3\). Tính \(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{3f\left(x\right)+10}+\sqrt[3]{f\left(x\right)-5}-2x-3}{x^2+x-6}\)
Em kiểm tra lại đề, chỗ \(f\left(x\right)-32\) kia có vẻ sai, vì như thế thì biểu thức đã cho ko phải dạng vô định
cho lim \(\dfrac{f\left(x\right)-5}{x-1}=4\) khi x->1 , lim \(\dfrac{g\left(x\right)-1}{x-1}=5\) khi x->1
tinh lim \(\dfrac{\sqrt{f\left(x\right)\times g\left(x\right)+4}-1}{x-1}\)khi x->1
Bạn tham khảo:
Nếu \(lim\) (x->1) \(\dfrac{f\left(x\right)-5}{x-1}=2\) và lim (x->1) \(\dfrac{g\left(x\right)-1}{x-1}=3\) thì lim (x->1... - Hoc24
Không giống hoàn toàn, nhưng cách làm thì giống hoàn toàn
Nếu \(lim\) (x->1) \(\dfrac{f\left(x\right)-5}{x-1}=2\) và lim (x->1) \(\dfrac{g\left(x\right)-1}{x-1}=3\) thì lim (x->1) \(\dfrac{\sqrt{f\left(x\right).g\left(x\right)+4}-3}{x-1}\) bằng mấy
Do \(x-1\rightarrow0\) khi \(x\rightarrow1\) nên \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-5}{x-1}=2\) hữu hạn khi và chỉ khi \(f\left(x\right)-5=0\) có nghiệm \(x=1\)
\(\Leftrightarrow f\left(1\right)-5=0\Rightarrow f\left(1\right)=5\)
Tương tự ta có \(g\left(1\right)=1\)
Do đó: \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{f\left(x\right).g\left(x\right)+4}-3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right).g\left(x\right)-5}{\left(x-1\right)\left(\sqrt{f\left(x\right).g\left(x\right)+4}+3\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\left[f\left(x\right)-5\right].g\left(x\right)+5\left[g\left(x\right)-1\right]}{\left(x-1\right)\left(\sqrt{f\left(x\right).g\left(x\right)+4}+3\right)}\)
\(=\left(2.1+5.3\right).\dfrac{1}{\sqrt{5.1+4}+3}=\dfrac{17}{6}\)
cho f(x) là 1 đa thức thoa man \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}=24\). tính \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{\left(x-1\right)\left(\sqrt{2f\left(x\right)+4}+6\right)}\)
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}\) hữu hạn nên \(f\left(x\right)-16=0\) có nghiệm \(x=1\)
\(\Rightarrow f\left(1\right)=16\)
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-16}{x-1}.\dfrac{1}{\sqrt{2f\left(x\right)+4}+6}=24.\dfrac{1}{\sqrt{2.16+4}+6}=2\)