Đáp án:3a)\(\frac{1}{6}\)
b)\(\frac{1}{2}\)
c)0
Giải thích các bước giải:
3a)\(\lim_{x\rightarrow 2}\frac{\sqrt{4x+1}-3}{x^{2}-4}\)
=\(\lim_{x\rightarrow 2}\frac{4x+1-9}{(x-2)(x+2)(\sqrt{4x+1}+3)}\)
=\(\lim_{x\rightarrow 2}\frac{4x-8}{(x-2)(x+2)(\sqrt{4x+1}+3)}\)
=\(\lim_{x\rightarrow 2}\frac{4(x-2)}{(x-2)(x+2)(\sqrt{4x+1}+3)}\)
=\(\lim_{x\rightarrow 2}\frac{4}{(x+2)(\sqrt{4x+1}+3)}\)
=\(\frac{1}{6}\)
b)\(\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{\sqrt[3]{4x+4}-2}\)
=\(\lim_{x\rightarrow 1}\frac{(x-1)(\sqrt[3]{4x+4}+2)}{4x+4-8.(\sqrt[3]{x}+1)}\)
=\(\lim_{x\rightarrow 1}\frac{(x-1)(\sqrt[3]{4x+4}+2)}{4(x-1)(\sqrt[3]{x}+1)}\)
=\(\lim_{x\rightarrow 1}\frac{\sqrt[3]{4x+4}+2}{4(\sqrt[3]{x}+1)}\)
=\(\frac{1}{2}\)
c)\(\lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1}{x}\)
=\(\lim_{x\rightarrow 0}\frac{1+x^{2}-1}{x(\sqrt{1+x^{2}}+1)}\)
=\(\lim_{x\rightarrow 0}\frac{x^{2}}{x(\sqrt{1+x^{2}}+1)}\)
=\(\lim_{x\rightarrow 0}\frac{x}{\sqrt{1+x^{2}}+1}\)
=\(\frac{0}{2}=0\)
câu 4 mình chưa học nên ko giải được
