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Question Number 183350 by Spillover last updated on 25/Dec/22

$$Find$$$$\left(a\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{3x+2}{x^2-x+1}$$$$\left(b\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x^2-1}}{2x+1}$$$$\left(c\right)\underset{x\to 5}{\lim }\frac{\sqrt{3x+1}-4}{x-5}$$

Commented by CElcedricjunior last updated on 25/Dec/22

$$\mathit{a})0$$$$\mathit{b})\frac{1}{2}$$$$\mathit{c})\frac{3}{8}$$$$$$

Commented by Spillover last updated on 25/Dec/22

$$thankyou$$

Answered by Spillover last updated on 25/Dec/22

$$$$$$\left(a\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{3x+2}{x^2-x+1}$$$$Divideby{x}^{2}upanddown$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{3}{x}+\frac{2}{x}}{1-\frac{1}{x}+\frac{1}{x^2}}x\to \mathrm{\infty }x=0$$$$\frac{0+0}{1-0+0}=0$$

Answered by Spillover last updated on 25/Dec/22

$$\left(b\right)\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x^2-1}}{2x+1}$$$$dividebyxupanddown$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x^2-1}}{2x+1}=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{\sqrt{x^2-1}}{x}}{\frac{2x+1}{x}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{1-\frac{1}{x^2}}}{2+\frac{1}{x}}$$$$x\to \mathrm{\infty }$$$$\frac{\sqrt{1-0}}{2+0}=\frac{1}{2}$$$$$$

Answered by Spillover last updated on 25/Dec/22

$$\left(c\right)\underset{x\to 5}{\lim }\frac{\sqrt{3x+1}-4}{x-5}$$$$Rationalizenumerator$$$$\underset{x\to 5}{\lim }\frac{\sqrt{3x+1}-4}{x-5}\times \frac{\sqrt{3x+1}+4}{\sqrt{3x+1}+4}$$$$\underset{x\to 5}{\lim }\frac{3x+1-16}{(x-5)(\sqrt{3x+1}+4)}$$$$\underset{x\to 5}{\lim }\frac{3}{\sqrt{3x+1}+4}$$$$x\to 5x=5$$$$\frac{3}{\sqrt{3\times 5+1}+4}=\frac{3}{8}$$$$$$

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