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Question Number 65779 by mathmax by abdo last updated on 03/Aug/19

$$calculate{lim}_{x\to 0}\frac{sin\left(x^2\right)-xtan\left(x\right)}{1-cos\left(4x\right)}$$

Commented by mathmax by abdo last updated on 04/Aug/19

$$letf\left(x\right)=sin\left(x^2\right)-xtanxandg\left(x\right)=1-cos\left(4x\right)$$$$wehave{lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}g\left(x\right)=0$$$$f^{{}^{\prime }}\left(x\right)=2xcos\left(x^2\right)-(tanx+x(1+{tan}^{2}x))\Rightarrow$$$${lim}_{x\to 0}{f}^{{}^{\prime }}\left(x\right)=0$$$$g^{{}^{\prime }}\left(x\right)=4sin\left(4x\right)\Rightarrow {lim}_{x\to 0}{g}^{{}^{\prime }}\left(x\right)=0$$$$f^{{}^{″}}\left(x\right)=2cos\left(x^2\right)-4x^{2}sin\left(x^2\right)-\{1+{tan}^{2}x+1+{tan}^{2}x+x\left(2tanx\right)(1+{tan}^{2}x)$$$$\Rightarrow {lim}_{x\to 0}{f}^{{}^{″}}\left(x\right)=0$$$$g^{{}^{″}}\left(x\right)=4cos\left(4x\right)\Rightarrow {lim}_{x\to 0}{g}^{\left(2\right)}\left(0\right)=4\Rightarrow$$$${lim}_{x\to 0}\frac{sin\left(x^2\right)-xtanx}{1-cos\left(4x\right)}=0$$$$$$

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