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Question Number 176710 by eka last updated on 25/Sep/22

Answered by a.lgnaoui last updated on 26/Sep/22

$$3.{lim}_{x\to -1}\frac{\sqrt{x^2+3}}{x^2-2}=\frac{\sqrt{4}}{-1}=-2anser\left(\mathbf{D}\right)$$$$4.\frac{x^2+2x-8}{x-2}=\frac{(x-2)(x+4)}{x-2}=x+4$$$$lim\frac{x^2+2x-8}{x-2}=2+4=6anser\left(\mathbf{D}\right)$$

Answered by Rasheed.Sindhi last updated on 25/Sep/22

$$1.\underset{x\to 2}{\lim }\frac{4x+1}{3-2x}$$$$\because \underset{x\to 2}{\lim }(3-2x)\ne 0$$$$\therefore \underset{x\to 2}{\lim }\frac{4x+1}{3-2x}=\frac{4\left(1\right)+1}{3-2\left(1\right)}=\frac{5}{-1}=-5$$$$2.\underset{x\to 1}{\lim }\frac{x^2-3x-4}{x^2-2x-3}=\frac{1^2-3\left(1\right)-4}{{\left(1\right)}^2-2\left(1\right)-3(\ne 0)}$$$$=\frac{-6}{-4}=\frac{3}{2}$$

Answered by Rasheed.Sindhi last updated on 25/Sep/22

$$3.\underset{x\to -1}{\lim }\frac{\sqrt{x^2+3}}{x^2-2}=\frac{\sqrt{{(-1)}^2+3}}{{(-1)}^2-2(\ne 0)}$$$$=\frac{2}{-1}=-2$$$$4.\underset{x\to 2}{\lim }\frac{x^2+2x-8}{x-2}=\underset{x\to 2}{\lim }\frac{\left(\cancel{x-2}\right)(x+4)}{\cancel{x-2}}$$$$=2+4=6$$

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