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Question Number 195365 by Mr.D.N. last updated on 31/Jul/23

      _(x→y) ^(lim)  ((tan x−tany)/(x−y))

$$\:\: \\ $$$$\:\underset{\mathrm{x}\rightarrow\mathrm{y}} {\overset{\mathrm{lim}} {\:}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{tany}}{\mathrm{x}−\mathrm{y}} \\ $$$$ \\ $$

Answered by cortano12 last updated on 31/Jul/23

  lim_(x→y)  ((tan (x−y)(1+tan x tan y))/(x−y))   = 1+tan^2 y = sec^2 y

$$\:\:\underset{{x}\rightarrow{y}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left({x}−{y}\right)\left(\mathrm{1}+\mathrm{tan}\:{x}\:\mathrm{tan}\:{y}\right)}{{x}−{y}} \\ $$$$\:=\:\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {y}\:=\:\mathrm{sec}\:^{\mathrm{2}} {y} \\ $$

Answered by kapoorshah last updated on 31/Jul/23

lim_(x→y)  ((sec^2 x )/1)        L′Hopital  = sec^2 y

$$\underset{{x}\rightarrow{y}} {\mathrm{lim}}\:\frac{\mathrm{sec}^{\mathrm{2}} {x}\:}{\mathrm{1}}\:\:\:\:\:\:\:\:{L}'{Hopital} \\ $$$$=\:\mathrm{sec}^{\mathrm{2}} {y} \\ $$

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