BeMath-Without-L-Hopital-and-series-find-lim-x-0-tan-x-x-x-3-

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

$$\:\spadesuit\mathcal{B}{e}\mathcal{M}{ath}\bigstar \\ $$$${Without}\:{L}'{Hopital}\:{and}\:{series}\: \\ $$$${find}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{3}} } \\ $$

Commented by bobhans last updated on 12/Aug/20

just comparing with L′Hopital lim_(x→0) ((sec^2 x−1)/(3x^2 )) = lim_(x→0) ((1−cos^2 x)/(3x^2 cos^2 x)) = lim_(x→0) ((sin^2 x)/(3x^2 )) × lim_(x→0) (1/(cos^2 x)) = (1/3)×1=(1/3)

$$\mathrm{just}\:\mathrm{comparing}\:\mathrm{with}\:\mathrm{L}'\mathrm{Hopital}\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}−\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{3x}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{3x}^{\mathrm{2}} }\:×\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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

Let x=2y L= lim_(x→0) ((tan x−x)/x^3 ) L= lim_(y→0) ((tan 2y−2y)/(8y^3 )) = lim_(y→0) ((((2tan y)/(1−tan^2 y))−2y)/(8y^3 )) = lim_(y→0) {((tan y−y (1−tan^2 y))/(4y^3 (1−tan^2 y)))} = lim_(y→0) {((tan y−y +ytan^2 y)/(4y^3 (1−tan^2 y)))} = lim_(y→0) {((tan y−y )/(4y^3 (1−tan^2 y)))}+lim_(y→0) {((ytan^2 y)/(4y^3 (1−tan^2 y)))} L = (L/4)+lim_(y→0) {((tan^2 y)/(4y^2 (1−tan^2 y)))} L = (L/4)+(1/4){lim_(y→0) (((tan^2 y)/(y^2 )))×lim_(y→0) (1/(1−tan^2 y))} ((3L)/4)=(1/4)×1 L=(1/3)

$${Let}\:{x}=\mathrm{2}{y} \\ $$$$\mathbb{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:{x}−{x}}{{x}^{\mathrm{3}} } \\ $$$$\mathbb{L}=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{2}{y}−\mathrm{2}{y}}{\mathrm{8}{y}^{\mathrm{3}} } \\ $$$$\:\:\:\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{\mathrm{2tan}\:{y}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}}−\mathrm{2}{y}}{\mathrm{8}{y}^{\mathrm{3}} } \\ $$$$\:\:\:\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{tan}\:{y}−{y}\:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)}{\mathrm{4}{y}^{\mathrm{3}} \:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)}\right\} \\ $$$$\:\:\:\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{tan}\:{y}−{y}\:+{y}\mathrm{tan}\:^{\mathrm{2}} {y}}{\mathrm{4}{y}^{\mathrm{3}} \:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)}\right\} \\ $$$$\:\:\:\:=\:\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{tan}\:{y}−{y}\:}{\mathrm{4}{y}^{\mathrm{3}} \:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)}\right\}+\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{{y}\mathrm{tan}\:^{\mathrm{2}} {y}}{\mathrm{4}{y}^{\mathrm{3}} \:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)}\right\} \\ $$$$\:\mathbb{L}\:\:\:=\:\frac{\mathbb{L}}{\mathrm{4}}+\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{tan}\:^{\mathrm{2}} {y}}{\mathrm{4}{y}^{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}\right)}\right\} \\ $$$$\:\mathbb{L}\:\:\:=\:\frac{\mathbb{L}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{tan}\:^{\mathrm{2}} {y}}{{y}^{\mathrm{2}} \:}\right)×\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {y}}\right\} \\ $$$$\frac{\mathrm{3}\mathbb{L}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{1} \\ $$$$\mathbb{L}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$

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

$${cooll}…{sir} \\ $$

Commented by bemath last updated on 11/Aug/20

$${thank}\:{you}\:{sir} \\ $$

Commented by malwaan last updated on 11/Aug/20

excuse me sir at the third line from down L=(L/4)+(1/4)lim_(y→0) {(((tan^2 y)/y^2 ))×lim_(y→0) (1/(1−tan^2 y))} ((3L)/4) = (1/4) × 1 L = (1/3) thanks

$${excuse}\:{me}\:{sir} \\ $$$${at}\:{the}\:{third}\:{line}\:{from}\:{down} \\ $$$$\mathbb{L}=\frac{\mathbb{L}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\left(\frac{{tan}^{\mathrm{2}} {y}}{{y}^{\mathrm{2}} }\right)×\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {y}}\right\} \\ $$$$\frac{\mathrm{3}\mathbb{L}}{\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:×\:\mathrm{1} \\ $$$$\mathbb{L}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${thanks} \\ $$

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

✔️

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

try to not imposing conditions for solving a question ...!

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

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

$${I}\:{disagree}. \\ $$$${It}\:{makes}\:{question}\:{more}\:{challanging}. \\ $$

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

there is no challenge in this forum there is no prize ...

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

Commented by bemath last updated on 12/Aug/20

$${sir}\:{Abdo},\:{why}\:{with}\:{my}\:{question}?\:{is}\:{there}\:{something}\:{wrong}\:?\:{or}\:{anyone}\:{object}\:{to}\:{that}\:{question}? \\ $$

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

No one is here for prizes.

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