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Question Number 31528 by abdo imad last updated on 09/Mar/18

$$find{lim}_{x\to 2}\frac{x^2-2^x}{arctanx-artan2}.$$

Commented by abdo imad last updated on 12/Mar/18

$$letputf\left(x\right)=2^x=e^{xln2}wehave{f}^{{}^{\prime }}\left(x\right)=ln\left(2\right).2^x$$$$f\left(x\right)=f\left(2\right)+\frac{x-2}{1!}{f}^{{}^{\prime }}\left(2\right)+\frac{{(x-2)}^2}{2!}\xi \left(x\right){with}_{x\to 2}\xi \left(x\right)\to 0$$$$f\left(x\right)=4+(x-2)ln2.2^2+\frac{{(x-2)}^2}{2!}\xi \left(x\right)\Rightarrow$$$$x^2-f\left(x\right)=x^2-4-ln\left(2\right)(x-2)2^2-\frac{{(x-2)}^2}{2!}\xi \left(x\right)$$$$=(x-2)(x+2-ln2.2^2-\frac{1}{2!}\xi \left(x\right))letputg\left(x\right)=arctanx$$$$wehaveg\left(x\right)=g\left(2\right)+\frac{x-2}{1!}{g}^{{}^{\prime }}\left(2\right)+\frac{{(x-2)}^2}{2!}\theta \left(x\right)with$$$$\theta {\left(x\right)}_{x\to 2}\to 0wehave{g}^{{}^{\prime }}\left(x\right)=\frac{1}{1+x^2}\Rightarrow g^{{}^{\prime }}\left(2\right)=\frac{1}{5}\Rightarrow$$$$g\left(x\right)=arctan2+\frac{1}{5}\frac{x-2}{1!}+\frac{{(x-2)}^2}{2!}\theta \left(x\right)\Rightarrow$$$$arctanx-arctan2=(x-2)(\frac{1}{5}+\frac{x-2}{2!}\theta \left(x\right))\Rightarrow$$$${lim}_{x\to 2}\frac{x^2-2^x}{arctanx-arctan2}$$$$={lim}_{x\to 2}\frac{x+2-4ln2-\frac{1}{2}\xi \left(x\right)}{\frac{1}{5}+\frac{x-2}{2!}\theta \left(x\right)}=5(4-4ln2).$$$$$$

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