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Question Number 113274 by bemath last updated on 12/Sep/20

$$\left(1\right)\underset{x\to 0}{\lim }\frac{\tan \mathrm{x}+4\tan 2\mathrm{x}-3\tan 3\mathrm{x}}{{\mathrm{x}}^2\tan \mathrm{x}}$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{\sqrt{\mathrm{x}}-\sqrt{\sin \mathrm{x}}}{{\mathrm{x}}^{\frac{3}{2}}}$$$$\left(3\right)\underset{x\to 0}{\lim }\frac{\sqrt{\mathrm{x}}+\sqrt{\sin \mathrm{x}}}{{\mathrm{x}}^{\frac{5}{2}}}$$

Commented by bemath last updated on 12/Sep/20

Commented by bobhans last updated on 13/Sep/20

$$\left(3\right)\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^2\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}+\sqrt{\sin \mathrm{x}})}=$$$$\underset{x\to 0}{\lim }\frac{\mathrm{x}-\sin \mathrm{x}}{{\mathrm{x}}^3}.\underset{x\to 0}{\lim }\frac{\mathrm{x}}{\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+\frac{{\mathrm{x}}^3}{6}})}$$$$=\frac{1}{6}.\underset{x\to 0}{\lim }\frac{\mathrm{x}}{\mathrm{x}(1+\sqrt{1+\frac{{\mathrm{x}}^2}{6}})}=\frac{1}{12}$$

Answered by john santu last updated on 12/Sep/20

$$\underset{x\to 0}{\lim }\frac{\tan x+4\tan 2x-3\tan 3x}{x^2\tan x}=L$$$$L=\underset{x\to 0}{\lim }\frac{x+\frac{x^3}{3}+4(2x+\frac{8x^3}{3})-3(3x+\frac{27x^3}{3})}{x^2(x+\frac{x^3}{3})}$$$$L=\underset{x\to 0}{\lim }\frac{11x^3-27x^3}{x^3(1+\frac{x^2}{3})}=-16.$$

Answered by Dwaipayan Shikari last updated on 12/Sep/20

$$\underset{x\to 0}{\lim }\frac{\sqrt{x}-\sqrt{x-\frac{x^3}{6}}}{x^{\frac{3}{2}}}=\frac{\sqrt{x}(1-\sqrt{1-\frac{x^2}{6}})}{x^{\frac{3}{2}}}=\frac{1-1+\frac{x^2}{12}}{x}=0$$

Answered by bemath last updated on 12/Sep/20

Answered by abdomsup last updated on 12/Sep/20

$$2)letf\left(x\right)=\frac{\sqrt{x}-\sqrt{sinx}}{x^{\frac{3}{2}}}\Rightarrow$$$$f\left(x\right)=\frac{1-x^{-\frac{1}{2}}\sqrt{sinx}}{x}$$$$wehavesinx\sim x-\frac{x^3}{6}\Rightarrow$$$$\sqrt{sinx}\sim \sqrt{x}\sqrt{1-\frac{x^2}{2}}\sim x^{\frac{1}{2}}(1-\frac{x^2}{4})\Rightarrow$$$$f\left(x\right)\sim \frac{1-(1-\frac{x^2}{4})}{x}=\frac{x}{4}\Rightarrow$$$${lim}_{x\to 0}f\left(x\right)=0$$$$$$

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