1) a) $\lim $ $\dfrac{n^3-2n+1}{2n^3-n+3}$
= $\lim $ $\dfrac{\dfrac{n^3}{n^3}-\dfrac{2n}{n^3}+\dfrac{1}{n^3}}{\dfrac{2n^3}{n^3}-\dfrac{n}{n^3}+\dfrac{3}{n^3}}$
= $\lim $ $\dfrac{1-\dfrac{2}{n^2}+\dfrac{1}{n^3}}{2-\dfrac{1}{n^2}+\dfrac{3}{n^3}}$
= $\dfrac{1-0+0}{2-0+0}$
= $\dfrac{1}{2}$
.
b) $\lim_{n \to 2} \dfrac{1-\sqrt[]{x^2-3}}{2-x}$
= $\lim_{n \to 2} \dfrac{(1-\sqrt[]{x^2-3})(1+\sqrt[]{x^2-3})}{(2-x)(1+\sqrt[]{x^2-3})}$
= $\lim_{n \to 2} \dfrac{1-x^2+3}{(2-x)(1+\sqrt[]{x^2-3})}$
= $\lim_{n \to 2} \dfrac{4-x^2}{(2-x)(1+\sqrt[]{x^2-3})}$
= $\lim_{n \to 2} \dfrac{(2-x)(2+x)}{(2-x)(1+\sqrt[]{x^2-3})}$
= $\lim_{n \to 2} \dfrac{2+x}{1+\sqrt[]{x^2-3}}$
= $\dfrac{2+2}{1+\sqrt[]{2^2-3}}$
= $\dfrac{4}{2}$ = $2$
.
c) $\lim_{n \to -\infty} \sqrt[]{x^2+x+3}+x$
= $\lim_{n \to -\infty} \dfrac{(\sqrt[]{x^2+x+3}+x)(\sqrt[]{x^2+x+3}-x)}{(\sqrt[]{x^2+x+3}-x)}$
= $\lim_{n \to -\infty} \dfrac{x^2+x+3-x^2}{(\sqrt[]{x^2+x+3}-x)}$
= $\lim_{n \to -\infty} \dfrac{x+3}{|x|(\sqrt[]{1+\dfrac{1}{x}+\dfrac{3}{x^2}}-x)}$
= $\lim_{n \to -\infty} \dfrac{x+3}{-x(\sqrt[]{1+\dfrac{1}{x}+\dfrac{3}{x^2}}-x)}$
= $\lim_{n \to -\infty} \dfrac{1+\dfrac{3}{x}}{-1(\sqrt[]{1+\dfrac{1}{x}+\dfrac{3}{x^2}}-1)}$
= $\dfrac{1+0}{-1(\sqrt[]{1+0+0}-1)}$
= $\dfrac{-1}{2}$
.
2) Ta có TXĐ: x # 0
Hàm số liên tục trên TXĐ
=> Hàm số liên tục trên 0
=> $\lim_{n \to 0^+} \dfrac{2-\sqrt[]{4-x}}{x}$ $= f(0)$
Ta có: $\lim_{n \to 0^+} \dfrac{2-\sqrt[]{4-x}}{x}$
= $\lim_{n \to 0^+} \dfrac{(2-\sqrt[]{4-x})(2+\sqrt[]{4-x})}{x(2+\sqrt[]{4-x})}$
= $\lim_{n \to 0^+} \dfrac{4-4+x}{x(2+\sqrt[]{4-x})}$
= $\lim_{n \to 0^+} \dfrac{x}{x(2+\sqrt[]{4-x})}$
= $\lim_{n \to 0^+} \dfrac{1}{2+\sqrt[]{4-x}}$
= $ \dfrac{1}{2+\sqrt[]{4-0}}$
= $ \dfrac{1}{4}$
Để hàm số liên tục
=> $f(0)$ =$\lim_{n \to 0^+} \dfrac{2-\sqrt[]{4-x}}{x}$= $ \dfrac{1}{4}$
<=> $2.0^2-5m+1= \dfrac{1}{4}$
<=> $-5m+1= \dfrac{1}{4}$
<=> $m= \dfrac{3}{20}$
