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Question Number 99720 by bobhans last updated on 23/Jun/20

lim_(x→0) (1/x^2 ) − (1/(tan^2 x)) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:? \\ $$

Commented by john santu last updated on 23/Jun/20

lim_(x→0) ((tan^2 x−x^2 )/(x^2 tan^2 x))= lim_(x→0) (((tan x+x)(tan x−x))/(x^2 tan^2 x)) = lim_(x→0) (((x+(x^3 /3)+x)(x+(x^3 /3)−x))/x^4 ) = lim_(x→0) ((x(2+(x^2 /3))((x^3 /3)))/x^4 ) = (2/3)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{tan}\:\mathrm{x}+\mathrm{x}\right)\left(\mathrm{tan}\:\mathrm{x}−\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{x}\right)\left(\mathrm{x}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}\left(\mathrm{2}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{3}}\right)\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right)}{\mathrm{x}^{\mathrm{4}} }\:=\:\frac{\mathrm{2}}{\mathrm{3}}\: \\ $$

Commented by mr W last updated on 23/Jun/20

=(1/x^2 )−((1/x^2 )−(2/3)+(x^2 /(15))−...) →(2/3)

$$=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{3}}+\frac{{x}^{\mathrm{2}} }{\mathrm{15}}−...\right) \\ $$$$\rightarrow\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented by john santu last updated on 23/Jun/20

joos...tooss prof W

$$\mathrm{joos}...\mathrm{tooss}\:\mathrm{prof}\:\mathrm{W} \\ $$

Commented by bobhans last updated on 24/Jun/20

sir w, how you get (1/x^2 )−(2/3)+(x^2 /(15))−... ?

$$\mathrm{sir}\:\mathrm{w},\:\mathrm{how}\:\mathrm{you}\:\mathrm{get}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{15}}−...\:?\: \\ $$

Commented by mr W last updated on 24/Jun/20

Maclaurin series of (1/(tan^2 x))

$${Maclaurin}\:{series}\:{of}\:\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:{x}} \\ $$

Answered by Ar Brandon last updated on 23/Jun/20

l=lim_(x→0) {(1/x^2 )−(1/(tan^2 x))}=lim_(x→0) {(1/x^2 )−((cos^2 x)/(sin^2 x))}=lim_(x→0) {((sin^2 x−x^2 cos^2 x)/(x^2 sin^2 x))} =lim_(x→0) {((2sinxcosx−(−2x^2 sinxcosx+2xcos^2 x))/(2x^2 sinxcosx+2xsin^2 x))} I tried this but obviously it isn′t the best approach. You may like to try Maclaurine′s expansion.

$${l}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \mathrm{x}}\right\}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\right\}=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}−\mathrm{x}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{x}}\right\} \\ $$$$\:\:=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{2sinxcosx}−\left(−\mathrm{2x}^{\mathrm{2}} \mathrm{sinxcosx}+\mathrm{2xcos}^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{2x}^{\mathrm{2}} \mathrm{sinxcosx}+\mathrm{2xsin}^{\mathrm{2}} \mathrm{x}}\right\} \\ $$$${I}\:{tried}\:{this}\:{but}\:{obviously}\:{it}\:{isn}'{t}\:{the}\:{best}\:{approach}. \\ $$$${You}\:{may}\:{like}\:{to}\:{try}\:{Maclaurine}'{s}\:{expansion}. \\ $$

Commented by bobhans last updated on 23/Jun/20

yes. L′hopital maybe alternative method

$$\mathrm{yes}.\:\mathrm{L}'\mathrm{hopital}\:\mathrm{maybe}\:\mathrm{alternative}\:\mathrm{method} \\ $$

Answered by john santu last updated on 23/Jun/20

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