The limit as x approaches zero of sin⁡x−xcos⁡xx3

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The limit as x approaches zero of  sin⁡x−xcos⁡xx3

rogreenhoxa8 · Accepted Answer

Let f(x)=g(x)h(x),where g(x)=sin⁡x−x.cos⁡x,andh(x)=x3.The limit is x→0.Since g(0) and h(0) both =0,then use l&#039;Hopital&#039;s rule, limf(x)[x−&amp;gt;0]=lim[x−&amp;gt;0]g″(x)/h″(x)g′(x)=cos⁡x−cos⁡x+x.sin⁡x=x.sin⁡xh′(x)=3x2Since g&#039;(0) and h&#039;(0) both = 0,then again use l&#039;Hopital&#039;s rule, i.e. limf(x)[x−&amp;gt;0]=lim[x−&amp;gt;0]g″(x)/h″(x)g″(x)=sin⁡x+x.cos⁡xh″(x)=6xSince g&#039;&#039;(0) and h&#039;&#039;(0) both = 0,then again use l&#039;Hopital&#039;s rule, i.e. limf(x)[x−&amp;gt;0]=lim[x−&amp;gt;0]g‴(x)/h‴(x)g‴(x)=cos⁡x+cos⁡x−x.sin⁡x=2cos⁡x−x.sin⁡xh‴(x)=6Now, g″(0)=2, and h‴(0)=6.So, limf(x)[x−&amp;gt;0]=lim[x−&amp;gt;0]g‴(x)/h‴(x)=lim[x−&amp;gt;0]2/6=1/3

The limit as x approaches zero of (sinx - xcosx)/x^3

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