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Question Number 51214 by gunawan last updated on 25/Dec/18

$$1.\underset{x\to -\frac{\mathrm{i}}{2}}{\lim }\frac{{(z-i)}^2}{(2z-i)(3-z)}$$$$2.\underset{x\to e^{\frac{\pi i}{4}}}{\lim }\frac{2z^2}{z^3-z-1}$$$$3.\underset{x\to 2i}{\lim }\frac{2z^2+8}{\sqrt{z^4}{-}^3\sqrt{64}}$$$$4.\underset{x\to 0}{\lim }\frac{\cos 4z-1}{z\sin z}$$$$$$

Answered by afachri last updated on 25/Dec/18

$$\left(4\right)\underset{z\to 0}{\lim }\frac{\left(\cos 4z\right)-1}{z\sin z}=\underset{z\to 0}{\lim }\frac{-{2\sin }^{2}2z}{z\sin z}$$$$=\underset{z\to 0}{\lim }\frac{-2\sin 2z.\sin 2z}{z\sin z}.\frac{2z.2z}{2z.2z}$$$$=\underset{z\to 0}{\lim }\frac{-2\left(2z\right)\left(2z\right)}{z\sin z}$$$$=-8$$

Answered by afachri last updated on 25/Dec/18

$$\left(3\right)\mathrm{know}:i=\sqrt{-1}$$$$i^2=-1$$$$i^4=1$$$$\underset{x\to 2i}{\lim }\frac{2x^2+8}{\sqrt{x^4}-\sqrt[3]{{64}^{}}}=\underset{x\to 2i}{\lim }\frac{2(x^2+4)}{\sqrt{x^4}-4}.\frac{\sqrt{x^4}+4}{\sqrt{x^4}+4}$$$$=\underset{x\to 2i}{\lim }\frac{2(x^2+4)(\sqrt{x^4}+4)}{x^4-16}$$$$=\underset{x\to 2i}{\lim }\frac{2(x^2+4)(\sqrt{x^4}+4)}{(x^2+4)(x^2-4)}$$$$=\underset{x\to 2i}{\lim }\frac{2(\sqrt{x^4}+4)}{x^2-4}$$$$=\underset{x\to 2i}{\lim }\frac{2(\sqrt{{\left(2i\right)}^{4^{}}}+4)}{{\left(2i\right)}^2-4}$$$$=\underset{x\to 2i}{\lim }\frac{2(\sqrt{16i^{4^{}}}+4)}{4i^2-4}=\frac{2\times 8}{-8}$$$$=-2$$$$$$

Commented by gunawan last updated on 25/Dec/18

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

Commented by afachri last updated on 25/Dec/18

$$\mathrm{ur}\mathrm{welcome},\mathrm{Sir}.$$

Answered by malwaan last updated on 25/Dec/18

$$\left(4\right){\lim }_{\mathrm{z}\to 0}\frac{(\cos 4\mathrm{z}-1)(\cos 4\mathrm{z}+1)}{\mathrm{z}\mathrm{sinz}(\cos 4\mathrm{z}+1)}$$$$=\mathrm{li}\underset{\mathrm{z}\to 0}{\mathrm{m}}\frac{{\cos }^{2}4\mathrm{z}-1}{\mathrm{z}\mathrm{sinz}(\cos 4\mathrm{z}+1)}$$$$=\mathrm{li}\underset{\mathrm{z}\to 0}{\mathrm{m}}\frac{-{\sin }^{2}4\mathrm{z}}{\mathrm{z}\mathrm{sinz}(\cos 4\mathrm{z}+1)}$$$$={\lim }_{\mathrm{z}\to 0}\frac{-2\left[2\mathrm{sinz}\mathrm{cosz}\right]\cos 2\mathrm{z}\times 2\left[2\mathrm{sinz}\mathrm{cosz}\right]\cos 2\mathrm{z}}{\mathrm{z}\mathrm{sinz}(\cos 4\mathrm{z}+1)}$$$$=\frac{-2\times 2\times 2\times 2}{2}=-8$$

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