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Help to find limit of this function: $\lim \limits_{x \to 3} \frac{sin(\sqrt{2x^2-3x-5} -\sqrt{1+x} )}{\ln(x-1)-\ln(x+1)+\ln2}$
I tried myself to solve this limit, but i unsure with answer.
$\frac{0}{0}$=$\lim \limits_{x \to 3} $$\frac{cos(sqrt{2x^2-3x-5}-sqrt{1+x})}{\frac{1}{x-1}-\frac{1}{x+1}}(\frac{4x-3}{2sqrt{2x^2-3x-5}}-\frac{1}{2\sqrt{1+x}}) =$ $\lim \limits_{x \to 3}\frac{2}{1-\frac{1}{4}}=8$

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  • $\begingroup$ The denominator is both finite and non-vanishing so you can just plug in the values - unless I am missing something. $\endgroup$ – Chinny84 Nov 25 '16 at 17:56
  • $\begingroup$ You say that you have tried, please post at least one of those tries. It would be good to know how much, or how little you know about how to approach this type of problem. Also, there might be a typeo in your question, since the last limit you state is 8/3 and not 8. There is no x in the function, so it is the constant function 8/3. A tip to conclude is to use taylor expansions of the functions sin and ln, or, as it seems in this case, just plug in the values. $\endgroup$ – Christopher.L Nov 25 '16 at 17:58
  • $\begingroup$ However, plugging in will give you denominator 0, sorry. That obviously does not work. $\endgroup$ – Christopher.L Nov 25 '16 at 18:09
  • $\begingroup$ So, again, use taylor expansions. It looks like one of those questions tailored (no pun intended) for that approach. $\endgroup$ – Christopher.L Nov 25 '16 at 18:14
  • $\begingroup$ @Christopher.L this is my way check it $\endgroup$ – Kairliev Alibi Nov 25 '16 at 18:26

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