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Question Number 94764 by Hamida last updated on 20/May/20

Answered by prakash jain last updated on 20/May/20

$$1\mathrm{a}\left(\mathrm{i}\right)$$$$\underset{x\to 3}{\lim }=1\left(\mathrm{top}\mathrm{point}\mathrm{of}\mathrm{semicircle}\right)$$$$1\mathrm{a}\left(\mathrm{ii}\right)$$$$\underset{x\to -1^{-}}{\lim }=0$$$$1\mathrm{a}\left(\mathrm{iii}\right)$$$$\underset{x\to 0^{+}}{\lim }=0$$$$\left(\mathrm{iv}\right)$$$$\mathrm{function}\mathrm{is}\mathrm{not}\mathrm{continous}\mathrm{at}f\left(0\right)$$$$\underset{x\to 0^{-}}{\lim }f\left(x\right)=0=\underset{x\to 0^{+}}{\lim }f\left(x\right)$$$$\mathrm{However}f\left(0\right)=2$$$$\left(\mathrm{v}\right)$$$$\underset{x\to 3^{-}}{\lim }f\left(3\right)=1=\underset{x\to 3^{+}}{\lim }f\left(x\right)$$$$\mathrm{To}\mathrm{makd}\mathrm{function}\mathrm{continuous}$$$$f\left(3\right)\mathrm{value}\mathrm{should}\mathrm{be}\mathrm{changed}\mathrm{to}1.$$

Commented by prakash jain last updated on 20/May/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{2x^2+4}-6}{x^2-2x-8}$$$$\underset{x\to 0}{\lim }\frac{\sqrt{2x^2+4}-6}{x^2-2x-8}\times \frac{\sqrt{2x^2+4}+6}{\sqrt{2x^2+4}+6}$$$$\underset{x\to 0}{\lim }\frac{2x^2+4-36}{(x-4)(x+2)}\times \frac{1}{\sqrt{2x^2+4}+6}$$$$\underset{x\to 0}{\lim }\frac{2x^2-32}{(x-4)(x+2)}\times \frac{1}{\sqrt{2x^2+4}+6}$$$$\underset{x\to 0}{\lim }\frac{2(x+4)(x-4)}{(x-4)(x+2)}\times \frac{1}{\sqrt{2x^2+4}+6}$$$$\underset{x\to 0}{\lim }\frac{2(x+4)}{(x+2)}\times \frac{1}{\sqrt{2x^2+4}+6}$$$$putx=4$$$$=\frac{2\times 8}{6}\times \frac{1}{12}=\frac{2}{9}$$

Commented by student work last updated on 21/May/20

$$\mathrm{ans}:\frac{1}{2}$$

Answered by mathmax by abdo last updated on 21/May/20

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{{2\mathrm{x}}^2+4}-6}{{\mathrm{x}}^2-2\mathrm{x}-8}\mathrm{we}\mathrm{have}{\mathrm{x}}^2-2\mathrm{x}-8={\mathrm{x}}^2-16-2\mathrm{x}+8$$$$=(\mathrm{x}-4)(\mathrm{x}+4)-2(\mathrm{x}-4)=(\mathrm{x}-4)(\mathrm{x}+4-2)=(\mathrm{x}-4)(\mathrm{x}+2)$$$${\lim }_{\mathrm{x}\to 4}\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{x}\to 4}\frac{\frac{{2\mathrm{x}}^2+4-36}{\sqrt{{2\mathrm{x}}^2+4}+6}}{(\mathrm{x}-4)(\mathrm{x}+2)}$$$$={\lim }_{\mathrm{x}\to 4}\frac{{2\mathrm{x}}^2-32}{(\mathrm{x}-4)(\mathrm{x}+2)(\sqrt{{2\mathrm{x}}^2+4}+6)}$$$$={\lim }_{\mathrm{x}\to 4}\frac{2(\mathrm{x}-4)(\mathrm{x}+4)}{(\mathrm{x}-4)(\mathrm{x}+2)(\sqrt{{2\mathrm{x}}^2+4}+6)}$$$$={\lim }_{\mathrm{x}\to 4}\frac{2\mathrm{x}+8}{(\mathrm{x}+2)(\sqrt{{2\mathrm{x}}^2+4}+6)}=\frac{16}{4\left(12\right)}=\frac{4\times 4}{4\times 3\times 4}=\frac{1}{3}$$

Commented by prakash jain last updated on 21/May/20

$$={\lim }_{\mathrm{x}\to 4}\frac{2\mathrm{x}+8}{(\mathrm{x}+2)(\sqrt{{2\mathrm{x}}^2+4}+6)}=\frac{16}{4\left(12\right)}=\frac{4\times 4}{4\times 3\times 4}=\frac{1}{3}$$$$\mathrm{Calculation}\mathrm{mistake}\mathrm{shown}\mathrm{in}\mathrm{red}$$$$x+2=6$$$$\mathrm{so}\mathrm{it}\mathrm{should}\mathrm{be}=\frac{16}{6\times 12}=\frac{2}{9}$$

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