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Question Number 7453 by ankit036 last updated on 30/Aug/16

Commented by Rasheed Soomro last updated on 30/Aug/16

(A)  1

$$\left(\mathrm{A}\right)\:\:\mathrm{1} \\ $$

Answered by afy1991 last updated on 30/Aug/16

c option is ri8.

$${c}\:{option}\:{is}\:{ri}\mathrm{8}. \\ $$

Commented by FilupSmith last updated on 30/Aug/16

unfortunatly you are incorect.  please see my answer

$$\mathrm{unfortunatly}\:\mathrm{you}\:\mathrm{are}\:\mathrm{incorect}. \\ $$$$\mathrm{please}\:\mathrm{see}\:\mathrm{my}\:\mathrm{answer} \\ $$

Answered by FilupSmith last updated on 30/Aug/16

lim_(x→0)  ((e^x −1)/x) = (0/0)  apply L′Hopital′s Law  ∴=lim_(x→0)  (((d/dx)(e^x −1))/((d/dx)(x)))  =lim_(x→0)  (e^x /1)  =1    ∴ lim_(x→0)  ((e^x −1)/x) = 1

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}\:=\:\frac{\mathrm{0}}{\mathrm{0}} \\ $$$${apply}\:{L}'{Hopital}'{s}\:{Law} \\ $$$$\therefore=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\frac{{d}}{{dx}}\left({e}^{{x}} −\mathrm{1}\right)}{\frac{{d}}{{dx}}\left({x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} }{\mathrm{1}} \\ $$$$=\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −\mathrm{1}}{{x}}\:=\:\mathrm{1} \\ $$

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