Bài tập nâng cao · Bài 2
Bài tập nâng cao Giới hạn của hàm số · Bài 2
Tính giới hạn
a. $\lim\limits_{x\to -2}\frac{x^{4}-16}{x^{3}+2x^{2}}$
b. $\lim\limits_{x\to -4}\frac{x^{2}+3x-4}{x^{2}+4x}$
c. $\lim\limits_{x\to 1}\frac{x^{3}-1}{x\left( x+5 \right)-6}$
d. $\lim\limits_{x\to -5}\frac{x^{2}+2x-15}{x+5}$
e. $\lim\limits_{x\to 1}\frac{x+x^{2}+...+x^{n}-n}{x-1}$
a. $\lim\limits_{x\to -2}\frac{x^{4}-16}{x^{3}+2x^{2}}$
b. $\lim\limits_{x\to -4}\frac{x^{2}+3x-4}{x^{2}+4x}$
c. $\lim\limits_{x\to 1}\frac{x^{3}-1}{x\left( x+5 \right)-6}$
d. $\lim\limits_{x\to -5}\frac{x^{2}+2x-15}{x+5}$
e. $\lim\limits_{x\to 1}\frac{x+x^{2}+...+x^{n}-n}{x-1}$
Xem lời giải
Lời giải
a. $\lim\limits_{x\to -2}\frac{x^{4}-16}{x^{3}+2x^{2}}=\lim\limits_{x\to -2}\frac{\left( x^{2}-4 \right)\left( x^{2}+4 \right)}{x^{2}\left( x+2 \right)}=\lim\limits_{x\to -2}\frac{\left( x-2 \right)\left( x^{2}+4 \right)}{x^{2}}=-8$
b. $\lim\limits_{x\to -4}\frac{x^{2}+3x-4}{x^{2}+4x}=\lim\limits_{x\to -4}\frac{\left( x-1 \right)\left( x+4 \right)}{x\left( x+4 \right)}=\lim\limits_{x\to -4}\frac{x-1}{x}=\frac{5}{4}$
c. $\lim\limits_{x\to 1}\frac{x^{3}-1}{x\left( x+5 \right)-6}=\lim\limits_{x\to 1}\frac{\left( x-1 \right)\left( x^{2}+x+1 \right)}{\left( x-1 \right)\left( x+6 \right)}=\lim\limits_{x\to 1}\frac{x^{2}+x+1}{x+6}=\frac{3}{7}$
d. $\lim\limits_{x\to -5}\frac{x^{2}+2x-15}{x+5}=\lim\limits_{x\to -5}\frac{\left( x+5 \right)\left( x-3 \right)}{x+5}=\lim\limits_{x\to -5}\left( x-3 \right)=-8$
e. $\lim\limits_{x\to 1}\frac{x+x^{2}+...+x^{n}-n}{x-1}=\lim\limits_{x\to 1}\frac{x-1+x^{2}-1+...+x^{n}-1}{x-1}$
$=\lim\limits_{x\to 1}\left( 1+\frac{x^{2}-1}{x-1}+...+\frac{x^{n}-1}{x-1} \right)=\lim\limits_{x\to 1}\left( 1+x+1+...+x^{n-1}+x^{n-2}+1 \right)$
$=\lim\limits_{x\to 1}\left( \underbrace{1+...+1}_{n}+\underbrace{x+...+x}_{n-1}+...+\underbrace{x^{n-2}+x^{n-2}}_{2}+x^{n-1} \right)=\left( n+n-1+n-2+...+1 \right)=\frac{n}{2}\left( n+1 \right)$
b. $\lim\limits_{x\to -4}\frac{x^{2}+3x-4}{x^{2}+4x}=\lim\limits_{x\to -4}\frac{\left( x-1 \right)\left( x+4 \right)}{x\left( x+4 \right)}=\lim\limits_{x\to -4}\frac{x-1}{x}=\frac{5}{4}$
c. $\lim\limits_{x\to 1}\frac{x^{3}-1}{x\left( x+5 \right)-6}=\lim\limits_{x\to 1}\frac{\left( x-1 \right)\left( x^{2}+x+1 \right)}{\left( x-1 \right)\left( x+6 \right)}=\lim\limits_{x\to 1}\frac{x^{2}+x+1}{x+6}=\frac{3}{7}$
d. $\lim\limits_{x\to -5}\frac{x^{2}+2x-15}{x+5}=\lim\limits_{x\to -5}\frac{\left( x+5 \right)\left( x-3 \right)}{x+5}=\lim\limits_{x\to -5}\left( x-3 \right)=-8$
e. $\lim\limits_{x\to 1}\frac{x+x^{2}+...+x^{n}-n}{x-1}=\lim\limits_{x\to 1}\frac{x-1+x^{2}-1+...+x^{n}-1}{x-1}$
$=\lim\limits_{x\to 1}\left( 1+\frac{x^{2}-1}{x-1}+...+\frac{x^{n}-1}{x-1} \right)=\lim\limits_{x\to 1}\left( 1+x+1+...+x^{n-1}+x^{n-2}+1 \right)$
$=\lim\limits_{x\to 1}\left( \underbrace{1+...+1}_{n}+\underbrace{x+...+x}_{n-1}+...+\underbrace{x^{n-2}+x^{n-2}}_{2}+x^{n-1} \right)=\left( n+n-1+n-2+...+1 \right)=\frac{n}{2}\left( n+1 \right)$