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Question Number 11364 by Joel576 last updated on 22/Mar/17

$$\underset{x\to 0}{\lim }\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Mar/17

$$\frac{\sqrt{1+tgx}-\sqrt{1+sinx}}{x^3}=\frac{(1+tgx)-(1+sinx)}{(\sqrt{1+tgx}+\sqrt{1+sinx}).x^3}=$$$$L=\underset{x\to 0}{\lim }\frac{tgx-sinx}{(\sqrt{1+tgx}+\sqrt{1+sinx})x^3}=$$$$\underset{x\to 0}{\lim }\frac{\frac{x^3}{2}}{(\sqrt{1+0}+\sqrt{1+0})x^3}=\frac{1}{4}.$$$$$$

Commented by Joel576 last updated on 22/Mar/17

$$\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{how}\mathrm{to}\mathrm{get}\frac{x^3}{2}$$

Commented by sm3l2996 last updated on 22/Mar/17

$$\mathrm{we}\mathrm{have}\mathrm{tg}\left(\mathrm{x}\right)\sim \mathrm{x}+\frac{{\mathrm{x}}^3}{3}\mathrm{and}\sin \left(\mathrm{x}\right)\sim \mathrm{x}-\frac{{\mathrm{x}}^3}{3!}$$$$\mathrm{so}\mathrm{tg}\left(\mathrm{x}\right)-\sin \left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^3}{6}$$$$\mathrm{tg}\left(\mathrm{x}\right)-\sin \left(\mathrm{x}\right)\sim \frac{{\mathrm{x}}^3}{2}$$

Commented by sandy_suhendra last updated on 22/Mar/17

$$\mathrm{I}\mathrm{try}\mathrm{to}\mathrm{solve}\mathrm{with}\mathrm{another}\mathrm{way}$$$$\mathrm{for}\mathrm{tanx}-\mathrm{sinx}$$$$=\frac{\mathrm{sinx}}{\mathrm{cosx}}-\mathrm{sinx}$$$$=\frac{\mathrm{sinx}-\mathrm{sinx}.\mathrm{cosx}}{\mathrm{cosx}}$$$$=\frac{\mathrm{sinx}(1-\mathrm{cosx})}{\mathrm{cosx}}$$$$=\frac{\mathrm{sinx}.{2\sin }^2\left(\frac{1}{2}\mathrm{x}\right)}{\mathrm{cosx}}$$$$\mathrm{thus}\frac{2\mathrm{sinx}.{\sin }^2\left(\frac{1}{2}\mathrm{x}\right)}{{\mathrm{x}}^3(\sqrt{1+\mathrm{tanx}}+\sqrt{1+\mathrm{sinx}})}$$$$=\frac{2\mathrm{sinx}}{\mathrm{x}}\times \frac{{\sin }^2\left(\frac{1}{2}\mathrm{x}\right)}{{\mathrm{x}}^2}\times \frac{1}{\sqrt{1+\mathrm{tanx}}+\sqrt{1+\mathrm{sinx}}}$$$$=2\times {\left(\frac{1}{2}\right)}^2\times \frac{1}{\sqrt{1+\tan 0}+\sqrt{1+\sin 0}}$$$$=\frac{1}{4}$$

Commented by Joel576 last updated on 23/Mar/17

$$thankyouverymuch$$

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