NIMCET 2024 Paper

Section: Mathematics

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Question : 3 of 120

Marks: +1, -0

The value of the limit

\lim ↙{x ⟶ 0} (\;{1^x+2^x+3^x+4^x}/{4}) ^{1 \/ x}

is:

1
$3^{1 \/ 3!}$
$3^{1 \/ 4}$
$4!^{1 \/ 4}$

Solution:

Given that

x ⟶ 0

, hence

1^x ≈ 2^x ≈ 3^x ≈ 4^x ≈ 1

Since all the numbers are tending to the same value, we can take their Geometric Mean in place of Arithmetic Mean.

(\;{1^x+2^x+3^x+4^x}/{4}) = (1^x 2^x 3^x 4^x) ^{1 \/ 4}

And the value of the limit

= [ (1^x 2^x 3^x 4^x) ^{1 \/ 4}] ^{1 \/ x}

=24^{1 \/ 4}=(4!)^{1 \/ 4}

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