Лимити функсияҳои зерро ҳисоб кунед:
1)\(\lim\limits_{n\to1}\frac{2n^2-3n-5}{n+1}\)
\(\lim\limits_{n\to1}\frac{2n^2-3n-5}{n+1}=\frac{2\cdot1^2-3\cdot1-5}{1+1}=\frac{2-3-5}{2}=\frac{-6}{2}=-3\).
Ҷавоб: \(\lim\limits_{n\to1}\frac{2n^2-3n-5}{n+1}=-3\).

2)\(\lim\limits_{x\to5}(3x-x^2)\)
\(\lim\limits_{x\to5}(3x-x^2)=3\cdot5-5^2=15-25=-10\).
Ҷавоб: \(\lim\limits_{x\to5}(3x-x^2)=-10\).

3)\(\lim\limits_{x\to1}(2x^2+3x-5)\)
\(\lim\limits_{x\to1}(2x^2+3x-5)=2\cdot1^2+3\cdot1-5=2+3-5=5-5=0\).
Ҷавоб: \(\lim\limits_{x\to1}(2x^2+3x-5)=0\).

4)\(\lim\limits_{x\to0}\frac{x^2-1}{2x^2-x-1}\)
\(\lim\limits_{x\to0}\frac{x^2-1}{2x^2-x-1}=\frac{0^2-1}{2\cdot0^2-0-1}=\frac{0-1}{0-0-1}=\frac{-1}{-1}=1\).
Ҷавоб: \(\lim\limits_{x\to0}\frac{x^2-1}{2x^2-x-1}=1\).

5)\(\lim\limits_{x\to\frac{1}{2}}\frac{2x^2+5x-3}{x^2-6x-7}\)
\(\lim\limits_{x\to\frac{1}{2}}\frac{2x^2+5x-3}{x^2-6x-7}=\frac{2\cdot(\frac{1}{2})^2+5\cdot\frac{1}{2}-3}{(\frac{1}{2})^2-6\cdot\frac{1}{2}-7}=\frac{2\cdot\frac{1}{4}+\frac{5}{2}-3}{\frac{1}{4}-3-7}=\frac{\frac{1}{2}+\frac{5}{2}-3}{\frac{1}{4}-10}=\frac{\frac{1+5}{2}-3}{-9\frac{3}{4}}=\frac{3-3}{-9\frac{3}{4}}=\frac{0}{-9\frac{3}{4}}=0\).
Ҷавоб: \(\lim\limits_{x\to\frac{1}{2}}\frac{2x^2+5x-3}{x^2-6x-7}=0\).