Limits

Question

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

Comments

Is it $1?$

---

yes , please explain .........

---

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.

---

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........??

Answer 1

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

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$

Comments

yeah i'm not good with calculus. can you post the correct answer?

---

@Utkarsh ,I think ur solution is correct..one another way is just divide numerator and denominator by n²  in both ( 2n²/3n²+1) and its power..and put n = ∞ ,it will become (2/3)⁰ =1

---

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