Bài tập nâng cao · Bài 17
Bài tập nâng cao Giới hạn của hàm số · Bài 17
Tìm các giới hạn sau:
a. $\lim\limits_{x\to 2}\frac{x^{2}-3x+2}{x^{2}+x-6}$
b. $\lim\limits_{x\to -2}\frac{x^{3}+3x^{2}+2x}{x^{2}-x-6}$
c. $\lim\limits_{x\to -1}\frac{x^{5}+1}{x^{3}+1}$
d. $\lim\limits_{x\to 2}\frac{x^{3}+3x^{2}-9x-2}{x^{3}-x-6}$
e. $\lim\limits_{x\to 1}\frac{x+x^{2}+...+x^{n}-n}{x-1}$
a. $\lim\limits_{x\to 2}\frac{x^{2}-3x+2}{x^{2}+x-6}$
b. $\lim\limits_{x\to -2}\frac{x^{3}+3x^{2}+2x}{x^{2}-x-6}$
c. $\lim\limits_{x\to -1}\frac{x^{5}+1}{x^{3}+1}$
d. $\lim\limits_{x\to 2}\frac{x^{3}+3x^{2}-9x-2}{x^{3}-x-6}$
e. $\lim\limits_{x\to 1}\frac{x+x^{2}+...+x^{n}-n}{x-1}$
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Lời giải
a. $\lim\limits_{x\to 2}\frac{x^{2}-3x+2}{x^{2}+x-6}=\lim\limits_{x\to 2}\frac{\left( x-2 \right)\left( x-1 \right)}{\left( x-2 \right)\left( x+3 \right)}=\lim\limits_{x\to 2}\frac{x-1}{x+3}=\frac{1}{5}$
b. $\lim\limits_{x\to -2}\frac{x^{3}+3x^{2}+2x}{x^{2}-x-6}=\lim\limits_{x\to -2}\frac{x\left( x+2 \right)\left( x+1 \right)}{\left( x+2 \right)\left( x-3 \right)}=-\frac{2}{5}$
c. $\lim\limits_{x\to -1}\frac{x^{5}+1}{x^{3}+1}=\lim\limits_{x\to -1}\frac{\left( x+1 \right)\left( x^{4}-x^{3}+x^{2}-x+1 \right)}{\left( x+1 \right)\left( x^{2}-x+1 \right)}=\frac{5}{3}$
d. $\lim\limits_{x\to 2}\frac{x^{3}+3x^{2}-9x-2}{x^{3}-x-6}=\lim\limits_{x\to 2}\frac{\left( x-2 \right)\left( x^{2}+5x+1 \right)}{\left( x-2 \right)\left( x^{2}+2x+3 \right)}=\frac{15}{11}$
e. $\lim\limits_{x\to 1}\frac{x-1+x^{2}-1+...+x^{n}-1}{x-1}=\lim\limits_{x\to 1}\underbrace{1+...+1}_{n\text{ số }1}+\underbrace{x+...+x}_{n-1\text{ số }x}+...+x^{n-1}=n+n-1+...+2+1=\frac{n\left( n+1 \right)}{2}$
b. $\lim\limits_{x\to -2}\frac{x^{3}+3x^{2}+2x}{x^{2}-x-6}=\lim\limits_{x\to -2}\frac{x\left( x+2 \right)\left( x+1 \right)}{\left( x+2 \right)\left( x-3 \right)}=-\frac{2}{5}$
c. $\lim\limits_{x\to -1}\frac{x^{5}+1}{x^{3}+1}=\lim\limits_{x\to -1}\frac{\left( x+1 \right)\left( x^{4}-x^{3}+x^{2}-x+1 \right)}{\left( x+1 \right)\left( x^{2}-x+1 \right)}=\frac{5}{3}$
d. $\lim\limits_{x\to 2}\frac{x^{3}+3x^{2}-9x-2}{x^{3}-x-6}=\lim\limits_{x\to 2}\frac{\left( x-2 \right)\left( x^{2}+5x+1 \right)}{\left( x-2 \right)\left( x^{2}+2x+3 \right)}=\frac{15}{11}$
e. $\lim\limits_{x\to 1}\frac{x-1+x^{2}-1+...+x^{n}-1}{x-1}=\lim\limits_{x\to 1}\underbrace{1+...+1}_{n\text{ số }1}+\underbrace{x+...+x}_{n-1\text{ số }x}+...+x^{n-1}=n+n-1+...+2+1=\frac{n\left( n+1 \right)}{2}$