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Question Number 107536 by bemath last updated on 11/Aug/20

$$\mathrm{♠}\mathcal{B}e\mathcal{M}ath★$$$$Without{L}^{\prime }Hopitalandseries$$$$find\underset{x\to 0}{\lim }\frac{\tan x-x}{x^3}$$

Commented by bobhans last updated on 12/Aug/20

$$\mathrm{just}\mathrm{comparing}\mathrm{with}{\mathrm{L}}^{\prime }\mathrm{Hopital}$$$$\underset{x\to 0}{\lim }\frac{\sec {}^2\mathrm{x}-1}{{3\mathrm{x}}^2}=\underset{x\to 0}{\lim }\frac{1-\cos {}^2\mathrm{x}}{{3\mathrm{x}}^2\cos {}^2\mathrm{x}}$$$$=\underset{x\to 0}{\lim }\frac{\sin {}^2\mathrm{x}}{{3\mathrm{x}}^2}\times \underset{x\to 0}{\lim }\frac{1}{\cos {}^2\mathrm{x}}=\frac{1}{3}\times 1=\frac{1}{3}$$

Answered by $@y@m last updated on 12/Aug/20

$$Letx=2y$$$$\mathbb{L}=\underset{x\to 0}{\lim }\frac{\tan x-x}{x^3}$$$$\mathbb{L}=\underset{y\to 0}{\lim }\frac{\tan 2y-2y}{8y^3}$$$$=\underset{y\to 0}{\lim }\frac{\frac{2\tan y}{1-\tan {}^{2}y}-2y}{8y^3}$$$$=\underset{y\to 0}{\lim }\left\{\frac{\tan y-y(1-\tan {}^{2}y)}{4y^3(1-\tan {}^{2}y)}\right\}$$$$=\underset{y\to 0}{\lim }\left\{\frac{\tan y-y+y\tan {}^{2}y}{4y^3(1-\tan {}^{2}y)}\right\}$$$$=\underset{y\to 0}{\lim }\left\{\frac{\tan y-y}{4y^3(1-\tan {}^{2}y)}\right\}+\underset{y\to 0}{\lim }\left\{\frac{y\tan {}^{2}y}{4y^3(1-\tan {}^{2}y)}\right\}$$$$\mathbb{L}=\frac{\mathbb{L}}{4}+\underset{y\to 0}{\lim }\left\{\frac{\tan {}^{2}y}{4y^2(1-\tan {}^{2}y)}\right\}$$$$\mathbb{L}=\frac{\mathbb{L}}{4}+\frac{1}{4}\left\{\underset{y\to 0}{\lim }\left(\frac{\tan {}^{2}y}{y^2}\right)\times \underset{y\to 0}{\lim }\frac{1}{1-\tan {}^{2}y}\right\}$$$$\frac{3\mathbb{L}}{4}=\frac{1}{4}\times 1$$$$\mathbb{L}=\frac{1}{3}$$

Commented by john santu last updated on 11/Aug/20

$$cooll...sir$$

Commented by bemath last updated on 11/Aug/20

$$thankyousir$$

Commented by malwaan last updated on 11/Aug/20

$$excusemesir$$$$atthethirdlinefromdown$$$$\mathbb{L}=\frac{\mathbb{L}}{4}+\frac{1}{4}\underset{y\to 0}{\lim }\left\{\left(\frac{{tan}^{2}y}{y^2}\right)\times \underset{y\to 0}{\lim }\frac{1}{1-{\tan }^{2}y}\right\}$$$$\frac{3\mathbb{L}}{4}=\frac{1}{4}\times 1$$$$\mathbb{L}=\frac{1}{3}$$$$thanks$$

Commented by $@y@m last updated on 12/Aug/20

✔️

Answered by mathmax by abdo last updated on 11/Aug/20

$$\mathrm{try}\mathrm{to}\mathrm{not}\mathrm{imposing}\mathrm{conditions}\mathrm{for}\mathrm{solving}\mathrm{a}\mathrm{question}...!$$

Commented by $@y@m last updated on 11/Aug/20

$$Idisagree.$$$$Itmakesquestionmorechallanging.$$

Commented by mathmax by abdo last updated on 11/Aug/20

$$\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{challenge}\mathrm{in}\mathrm{this}\mathrm{forum}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{prize}...$$

Commented by bemath last updated on 12/Aug/20

$$sirAbdo,whywithmyquestion?istheresomethingwrong?oranyoneobjecttothatquestion?$$

Commented by $@y@m last updated on 12/Aug/20

No one is here for prizes.

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