# Doubt from functions

Give detailed solution

apply formula \lim_{n\to\infty} e^{(f(x)-1){g(x)}}

Yes I have used to same by couldn’t able to simplify further @Sneha_2021...
Given ans: 1,3,4

can u share ur calculation i m try to manipulate solution otherwise will take time for writing calculation.. @Bhuvanitha_2020

Well, there is a neat little trick you can use here. Its not exactly a trick, but a sort of approximation to evaluate limits.

As x tends to zero,cosec x becomes 1/x, cosx becomes 1, sinx becomes x

so the limit tending to zero evaluates to:

((e^(1) - 2^(x^2))/(e-1))^(k/x)
= ((e-2^(x^2))/(e-1))^(k/x))
as x tends to zero,

= ((e-2^(0))/(e-1))^(k))
=1^k
= 1

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Sir but k/x tends to infinity...? 1^(k/x)

Use binomial expansion