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I am attempting to evaluate the limit:

$$\lim_\limits{x\to3}\frac{\sin(x+1)}{2x(x-3)}$$

I have stared at this limit for 20 min and cant seem to find an angle to attack it, was wondering if anyone can point me in the right direction to solving this limit.

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Since $\pi < 4 < \dfrac 32\pi$ we have that

  • $\sin (x+1) \to \sin 4 <0$

and then

$$\lim_\limits{x\to3^+}\frac{\sin(x+1)}{2x(x-3)}=-\infty$$

$$\lim_\limits{x\to3^-}\frac{\sin(x+1)}{2x(x-3)}=\infty$$

therefore the limit doesn't exist.

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  • $\begingroup$ what do you mean by sin 4 < 0? $\endgroup$ – Not Friedrich gauss Sep 5 '18 at 14:30
  • $\begingroup$ @user196104 4 radians are equal to about 230° and therefore $\sin 4 <0$. $\endgroup$ – gimusi Sep 5 '18 at 14:31
  • $\begingroup$ @user196104 or $\pi < 4 < \dfrac 32\pi$ implies $4$ radians is in the third quadrant implies $\sin 4 < 0$ $\endgroup$ – steven gregory Sep 5 '18 at 14:33
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    $\begingroup$ All that really matters is that $\sin4\neq0$. $\endgroup$ – mr_e_man Sep 5 '18 at 14:34
  • $\begingroup$ @stevengregory in that way it is more clear I update that! $\endgroup$ – gimusi Sep 5 '18 at 14:35

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