87 views   Find the value of above limit.

How to approach with  this type of infinity power infinity problems ??

im thinking like first we should convert this to 1 power infinity problem and then solve??

is it correct ?? please solve above que

| 87 views
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Is it $1?$
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+1
Simply take $t = \frac{1}{n}$ Now when $n \rightarrow \infty \text{ (approaches to)}$ then $t \rightarrow 0 \text{ (approaches to)}$

And put in the above expression. Then plugin t = 0. You get $({\frac{2}{3}})^0$ which is 1.
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Now when n→0 (approaches to)n→0 (approaches to) then t→0 (approaches to)

this is wrong i think

when n->0 then t-> to infinity........??

+1 vote
$\large \lim_{n\rightarrow \infty} \left ( \frac{2n^{2}}{3n^{2}+1} \right )^{\LARGE \frac{-3n+{2}}{5n^{2}-3}}$

Let n = $\large \frac{1}{t}$

$\large \lim_{t\rightarrow 0} \left ( \frac{\frac {2}{t^{2}}}{\frac {3}{t^{2}}+1} \right )^{\LARGE \frac{\frac {-3}{t} +{2}}{\frac {5}{t^{2}}-3}}$

$\large \lim_{t\rightarrow 0} \left ( \frac{2}{3 + t^{2}} \right )^{\LARGE \frac{t(-3 +{2t})}{5-3t^{2}}}$

$\large \left ( \frac{2}{3 + 0^{2}} \right )^0$

The answer is $1$
by Boss (36.5k points)
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yeah i'm not good with calculus. can you post the correct answer?
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@Utkarsh ,I think ur solution is correct..one another way is just divide numerator and denominator by nin both ( 2n2/3n2+1) and its power..and put n = ∞ ,it will become (2/3)0 =1

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ankitgupta yeah that's easier thanks i'll update the answer