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Question Number 137255 by Nith last updated on 31/Mar/21

Answered by Dwaipayan Shikari last updated on 31/Mar/21

$$sinx\sim x$$$$2^{sinx}\sim 2^x\sim 1+xlog\left(2\right)$$$$3^{sinx}\sim 1+xlog\left(3\right)$$$$\underset{x\to 0}{\lim }\frac{2^{sinx}-3^{sinx}}{x}=\frac{1+xlog\left(2\right)-1-xlog\left(3\right)}{x}=log\left(\frac{2}{3}\right)$$

Answered by JoseLuisEsponizaCasares last updated on 31/Mar/21

$$L^{\prime }{Hopital}^{\prime }sRule$$$$if\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{0}{0}then$$$$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{f{}^{\prime }\left(x\right)}{g{}^{\prime }\left(x\right)}$$$$f{}^{\prime }\left(x\right)=2^{sinx}cosxln2-3^{sinx}cosxln3$$$$f{}^{\prime }\left(0\right)=2^0\left(1\right)ln2-3^0\left(1\right)ln3$$$$=\left(1\right)\left(1\right)ln2-\left(1\right)\left(1\right)ln3=ln2-ln3$$$$=ln\left(\frac{2}{3}\right)$$$$g{}^{\prime }\left(x\right)=1$$$$g^{\prime }\left(0\right)=1$$$$Result$$$$\underset{x\to 0}{lim}\frac{f{}^{\prime }\left(x\right)}{g{}^{\prime }\left(x\right)}=\frac{ln\left(\frac{2}{3}\right)}{1}=ln\left(\frac{2}{3}\right)s$$

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