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Question Number 153154 by liberty last updated on 05/Sep/21
  lim_(x→y)  ((sin (e^x )−sin (e^y ))/(x−y))=?
$$\:\:\underset{{x}\rightarrow{y}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({e}^{{x}} \right)−\mathrm{sin}\:\left({e}^{{y}} \right)}{{x}−{y}}=? \\ $$
Answered by bramlexs22 last updated on 05/Sep/21
  lim_(x→y)  ((sin (e^x )−sin (e^y ))/(x−y)) =lim_(x→y) ((2cos (((e^x +e^y )/2))sin (((e^x −e^y )/2)))/(x−y))  =2cos (((2e^y )/2)).lim_(x→y) ((sin (((e^x −e^y )/2)))/(x−y))  =2cos (e^y ).lim_(t→0) ((sin (((e^(t+y) −e^y )/2)))/t)  =2cos (e^y ).lim_(t→0) ((sin ((1/2)e^y (e^t −1)))/t)  =2cos (e^y ).lim_(t→0)  ((e^y (e^t −1))/(2t))=2cos (e^y )((e^y /2))  =e^y  cos (e^y )
$$\:\:\underset{{x}\rightarrow\mathrm{y}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left(\mathrm{e}^{\mathrm{x}} \right)−\mathrm{sin}\:\left(\mathrm{e}^{\mathrm{y}} \right)}{\mathrm{x}−\mathrm{y}}\:=\underset{{x}\rightarrow\mathrm{y}} {\mathrm{lim}}\frac{\mathrm{2cos}\:\left(\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{\mathrm{y}} }{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{\mathrm{y}} }{\mathrm{2}}\right)}{\mathrm{x}−\mathrm{y}} \\ $$$$=\mathrm{2cos}\:\left(\frac{\mathrm{2e}^{\mathrm{y}} }{\mathrm{2}}\right).\underset{{x}\rightarrow\mathrm{y}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{\mathrm{y}} }{\mathrm{2}}\right)}{\mathrm{x}−\mathrm{y}} \\ $$$$=\mathrm{2cos}\:\left(\mathrm{e}^{\mathrm{y}} \right).\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\mathrm{e}^{\mathrm{t}+\mathrm{y}} −\mathrm{e}^{\mathrm{y}} }{\mathrm{2}}\right)}{\mathrm{t}} \\ $$$$=\mathrm{2cos}\:\left(\mathrm{e}^{\mathrm{y}} \right).\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{\mathrm{y}} \left(\mathrm{e}^{\mathrm{t}} −\mathrm{1}\right)\right)}{\mathrm{t}} \\ $$$$=\mathrm{2cos}\:\left(\mathrm{e}^{\mathrm{y}} \right).\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{y}} \left(\mathrm{e}^{\mathrm{t}} −\mathrm{1}\right)}{\mathrm{2t}}=\mathrm{2cos}\:\left(\mathrm{e}^{\mathrm{y}} \right)\left(\frac{\mathrm{e}^{\mathrm{y}} }{\mathrm{2}}\right) \\ $$$$=\mathrm{e}^{\mathrm{y}} \:\mathrm{cos}\:\left(\mathrm{e}^{\mathrm{y}} \right)\: \\ $$

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