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Question Number 131079 by mathlove last updated on 01/Feb/21

lim_(h→0) ((e^(2(√2)x) e^(2(√2)h) −e^(2(√2)x) )/h)=?

$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}{x}} {e}^{\mathrm{2}\sqrt{\mathrm{2}}{h}} −{e}^{\mathrm{2}\sqrt{\mathrm{2}}{x}} }{{h}}=? \\ $$

Commented by EDWIN88 last updated on 01/Feb/21

lim_(h→0) ((e^(2(√2) (x+h)) −e^(2(√2) x) )/h) = f ′(x) where f(x) = e^(2(√2) x) let y = e^(2(√2) x) ⇒ln (y)= 2(√2) x ((y′)/y) = 2(√2) ⇒y′ = 2(√2) e^(2(√2) x) so lim_(h→0) ((e^(2(√2) (x+h)) −e^(2(√2) x) )/h) = 2(√2) e^(2(√2) x)

$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:\left({x}+{h}\right)} −{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} }{{h}}\:=\:{f}\:'\left({x}\right) \\ $$$${where}\:{f}\left({x}\right)\:=\:{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} \\ $$$$\:{let}\:{y}\:=\:{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} \:\Rightarrow\mathrm{ln}\:\left({y}\right)=\:\mathrm{2}\sqrt{\mathrm{2}}\:{x} \\ $$$$\:\frac{{y}'}{{y}}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:\Rightarrow{y}'\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} \\ $$$${so}\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:\left({x}+{h}\right)} −{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} }{{h}}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} \\ $$$$ \\ $$

Answered by EDWIN88 last updated on 01/Feb/21

e^(2(√2) x) × lim_(h→0 ) ((e^(2(√2) h) −1)/h)

$${e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{x}} \:×\:\underset{{h}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}\:{h}} −\mathrm{1}}{{h}} \\ $$

Commented by mathlove last updated on 01/Feb/21

and lim_(h→0) ((e^(2(√2)h) −1)/h)=?

$${and} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}{h}} −\mathrm{1}}{{h}}=? \\ $$

Commented by Dwaipayan Shikari last updated on 01/Feb/21

lim_(x→0) ((e^(2(√2)x) −1)/x)=((1+2(√2)x−1)/x)=2(√2) lim_(x→0) e^x =1+x

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\mathrm{2}\sqrt{\mathrm{2}}{x}} −\mathrm{1}}{{x}}=\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}{x}−\mathrm{1}}{{x}}=\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{e}^{{x}} =\mathrm{1}+{x} \\ $$

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