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I tried to evaluate the following limits but I just couldn't succeed, basically I can't use L'Hopital to solve this...

for the second limit I tried to transform it into $e^{\frac{2n\sqrt{n+3}ln(\frac{3n-1}{2n+3})}{(n+4)\sqrt{n+1}}}$ but still with no success...

$$\lim_{n \to \infty } \frac{2n^2-3}{-n^2+7}\frac{3^n-2^{n-1}}{3^{n+2}+2^n}$$

$$\lim_{n \to \infty } \frac{3n-1}{2n+3}^{\frac{2n\sqrt{n+3}}{(n+4)\sqrt{n+1}}}$$

Any suggestions/help? :)

Thanks

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For the first limit, it breaks into 2 factors with finite limits.
$$\lim{n \to \infty} \frac{2n^2-3}{7-n^2} = \frac{2n^2}{-n^2} =-2\\ \lim{n \to \infty} \frac{3^n-2^{n-1}}{3^{n+2}+2^n2} = \frac{3^n}{3^{n+2}} = \frac{1}{9} $$ so the answer is $-\frac{2}{9}$.

For the second, rewrite it as $$ \left(\frac{(3n-1)(2n-3)}{4n^2-9} \right) ^{\frac{\sqrt{n}2n(1+\frac{3}{2n}+\ldots)}{\sqrt{n}(n+4)(1+\frac{1}{2n}+\ldots)}} $$ and expand to next-lowest order in $1/n$ to get $$ \left( \frac{3}{2} \left[ 1-\frac{11}{6n}+\ldots\right] \right)^{2(1+\frac{3}{2n}+\ldots-\frac{9}{2n}+\ldots)} $$ Since the exponent does not go to infinity we can in fact just use the lowest order terms, getting $$\left( \frac{3}{2} \right)^2 = \frac{9}{4}$$

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  • $\begingroup$ Really bad form to use $=$ that way. In particular, there is no $n$ variable in the left hand side. $\endgroup$ – Thomas Andrews Mar 19 '15 at 15:26
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Hints : $$ \frac{2n^2-3}{-n^2+7} = \frac{2 - \frac{3}{n^2}}{-1+\frac{7}{n^2}},$$ and $$ \frac{3^n-2^{n-1}}{3^{n+2}+2^n} = \frac{1-\frac{1}{2}\left( \frac{2}{3} \right)^n}{3^2+\left( \frac{2}{3} \right)^n}.$$

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For the second one pay attention to the order of the numerator and denominator: the largest terms converge to some constant, the rest to 0, so you should get $(\frac{3}{2})^2$.

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  • $\begingroup$ I think that it could be $(\frac{3}{2})^2$. If I am right, you have a typo. What do you think ? $\endgroup$ – Claude Leibovici Mar 19 '15 at 15:05
  • $\begingroup$ OK fixed, thanks. $\endgroup$ – Alex Mar 19 '15 at 15:07

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