Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 188933 by horsebrand11 last updated on 09/Mar/23

$$\underset{x\to 1}{\lim }\frac{\sqrt{4x-3}+\sqrt{2x-1}-3x+1}{x^2-2x+1}=?$$

Commented by TUN last updated on 09/Mar/23

$$$$$$=\underset{x\to 1}{\lim }\frac{\sqrt{4x-3}-(2x-1)+\sqrt{2x-1}-x}{{(x-1)}^2}$$$$=\underset{x\to 1}{\lim }\frac{-4}{\sqrt{4x-3}+2x-1}+\underset{x\to 1}{\lim }\frac{-1}{\sqrt{2x-1}+x}$$$$=\underset{x\to 1}{\lim }\frac{-4}{\sqrt{1}+1}+\underset{x\to 1}{\lim }\frac{-1}{1+1}$$$$=\underset{x\to 0}{\lim }-2+\frac{-1}{2}=-\frac{5}{2}$$$$$$

Answered by floor(10²Eta[1]) last updated on 09/Mar/23

$$\underset{x\to 1}{\lim }\frac{\frac{2}{\sqrt{4\mathrm{x}-3}}+\frac{1}{\sqrt{2\mathrm{x}-1}}-3}{2\mathrm{x}-2}$$$$\underset{x\to 1}{\lim }-\frac{\frac{4}{\sqrt{{(4\mathrm{x}-3)}^3}}+\frac{1}{\sqrt{{(2\mathrm{x}-1)}^3}}}{2}=\frac{5}{2}$$

Answered by cortano12 last updated on 09/Mar/23

$$\mathrm{L}=\underset{x\to 1}{\lim }\frac{\sqrt{4\mathrm{x}-3}+\sqrt{2\mathrm{x}-1}-3\mathrm{x}+1}{{\mathrm{x}}^2-2\mathrm{x}+1}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{\sqrt{4(\mathrm{x}+1)-3}+\sqrt{2(\mathrm{x}+1)-1}-3(\mathrm{x}+1)+1}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{\sqrt{4\mathrm{x}+1}+\sqrt{2\mathrm{x}+1}-3\mathrm{x}-2}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{(1+\frac{1}{2}\left(4\mathrm{x}\right)-\frac{1}{8}{\left(4\mathrm{x}\right)}^2)+(1+\frac{1}{2}\left(2\mathrm{x}\right)-\frac{1}{8}{\left(2\mathrm{x}\right)}^2)-3\mathrm{x}-2}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{1+2\mathrm{x}-{2\mathrm{x}}^2+1+\mathrm{x}-\frac{1}{2}{\mathrm{x}}^2-3\mathrm{x}-2}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{-\frac{5}{2}{\mathrm{x}}^2}{{\mathrm{x}}^2}=-\frac{5}{2}$$

Commented by leandrosriv02 last updated on 11/Mar/23

$$$$$$\mathrm{L}=\underset{x\to 1}{\lim }\frac{\sqrt{4\mathrm{x}-3}+\sqrt{2\mathrm{x}-1}-3\mathrm{x}+1}{{\mathrm{x}}^2-2\mathrm{x}+1}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{\sqrt{4(\mathrm{x}+1)-3}+\sqrt{2(\mathrm{x}+1)-1}-3(\mathrm{x}+1)+1}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{\sqrt{4\mathrm{x}+1}+\sqrt{2\mathrm{x}+1}-3\mathrm{x}-2}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{(1+\frac{1}{2}\left(4\mathrm{x}\right)-\frac{1}{8}{\left(4\mathrm{x}\right)}^2)+(1+\frac{1}{2}\left(2\mathrm{x}\right)-\frac{1}{8}{\left(2\mathrm{x}\right)}^2)-3\mathrm{x}-2}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{1+2\mathrm{x}-{2\mathrm{x}}^2+1+\mathrm{x}-\frac{1}{2}{\mathrm{x}}^2-3\mathrm{x}-2}{{\mathrm{x}}^2}$$$$\mathrm{L}=\underset{x\to 0}{\lim }\frac{-\frac{5}{2}{\mathrm{x}}^2}{{\mathrm{x}}^2}=-\frac{5}{2}$$$$$$$$$$$$\underset{x\to 1}{\lim }\frac{\sqrt{4\mathrm{x}-3}+\sqrt{2\mathrm{x}-1}-3\mathrm{x}+1}{{\mathrm{x}}^2-2\mathrm{x}+1}=-\frac{5}{2}$$$$\overline{)\begin{array}{ccccccc}x& 1.1& 1.01& 1.001& 0.999& 0.99& 0.9\\ f\left(x\right)& -2.1& -2.45& -2.495& -2.504& -2.54& -3.0\end{array}}$$$$\underset{x\to 1^{-}}{\lim }f\left(x\right)=-\frac{5}{2}$$$$\underset{x\to 1^{+}}{\lim }f\left(x\right)=-\frac{5}{2}$$$$Ellimiteexisteyesdiscontinuaevitableenx=1$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com