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Question Number 134006 by mohammad17 last updated on 26/Feb/21

Answered by malwan last updated on 26/Feb/21

$$\left(1\right)\underset{x\to 4}{lim}\frac{2x-5}{3}=\frac{2\times 4-5}{3}=1$$$$\left(2\right)\underset{x\to -3^{+}}{lim}\frac{\sqrt{x+4}-\sqrt{-x-2}}{\sqrt{x+3}}\times \frac{\sqrt{x+4}+\sqrt{-x-2}}{\sqrt{x+4}+\sqrt{-x-2}}$$$$=\underset{x\to -3^{+}}{lim}\frac{2\sqrt{x+3}\sqrt{x+3}}{\sqrt{x+3}(\sqrt{x+4}+\sqrt{-x-2})}=0$$$$but\underset{x\to -3^{-}}{lim}{doesn}^{\prime }texist$$$$\left(3\right)\underset{x\to \sqrt{2}}{lim}\frac{2-x^2}{\sqrt{2}-x}=\underset{x\to \sqrt{2}}{lim}\frac{(\sqrt{2}+x)(\sqrt{2}-x)}{\sqrt{2}-x}=2\sqrt{2}$$$$\left(4\right)\underset{x\to 0}{lim}\frac{sinx-cosxsinx}{x^2}$$$$=\underset{x\to 0}{lim}\frac{sinx(1-cosx)}{x^2}=\underset{x\to 0}{lim}\frac{sinx\times 2{sin}^2\frac{x}{2}}{x^2}$$$$=0$$$$\left(5\right)\underset{x\to 1}{lim}\frac{\frac{1}{\sqrt{x}}-1}{1-x}=\underset{x\to 1}{lim}\frac{\frac{1-\sqrt{x}}{\sqrt{x}}}{(1+\sqrt{x})(1-\sqrt{x})}$$$$=\frac{1}{2}$$$$\left(6\right)\underset{x\to \mathrm{\infty }}{lim}\frac{-x^7+3x^5+4}{7x^7-6x^6+5x^5+3}=-\frac{1}{7}$$

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