Question

\(a,\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-\sqrt{x^2+x+1}}{x^2-2x+1}\)

\(b,\lim\limits_{x\rightarrow7}\dfrac{\sqrt{x-3}-2}{49-x^2}\)

Answer 1

Lời giải:
a.

\(\lim\limits_{x\to 0}\frac{\sqrt{x+1}-\sqrt{x^2+x+1}}{x^2-2x+1}=\lim\limits_{x\to 0}\frac{\sqrt{0+1}-\sqrt{0^2+0+1}}{0^2-2.0+1}=0\)

b.

\(\lim\limits_{x\to 7}\frac{\sqrt{x-3}-2}{49-x^2}=\lim\limits_{x\to 7}\frac{(x-3)-2^2}{(49-x^2)(\sqrt{x-3}+2)}\)

\(=\lim\limits_{x\to 7}\frac{x-7}{-(x-7)(x+7)(\sqrt{x-3}+2)}=\lim\limits_{x\to 7}\frac{1}{-(x+7)(\sqrt{x-3}+2)}=\frac{1}{-(7+7)(\sqrt{7-3}+2)}=\frac{-1}{56}\)

 

Answer 2

a) \(\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {x + 1} - \sqrt {{x^2} + x + 1} }}{{{x^2} - 2x + 1}} = \frac{{\sqrt {0 + 1} - \sqrt {{0^2} + 0 + 1} }}{{{0^2} - 2.0 + 1}} = 0\)

b) \(\mathop {\lim }\limits_{x \to 7} \frac{{\sqrt {x - 3} - 2}}{{49 - {x^2}}} = \mathop {\lim }\limits_{x \to 7} \frac{{x - 3 - {2^2}}}{{\left( {7 - x} \right)\left( {7 + x} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \mathop {\lim }\limits_{x \to 7} \frac{{ - 1}}{{\left( {7 + x} \right)\left( {\sqrt {x - 3} + 2} \right)}} = \frac{{ - 1}}{{56}}\)