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NIMCET 2024 Mathematics PYQ — The value of the limit is… | Mathem Solvex | Mathem Solvex

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NIMCET 2024 — Mathematics PYQ

NIMCET | Mathematics | 2024

The value of the limit

$lim_{x \to 0} (\frac{1 ^{x} + 2 ^{x} + 3 ^{x} + 4 ^{x}}{4})^{1/ x}$ is

Choose the correct answer:

A.
$1$
B.
$3!^{1/3!}$
C.
$3!^{1/4}$
D.
$4!^{1/4}$
(Correct Answer)

Correct Answer:

$4!^{1/4}$

Explanation

Solution:

Let $L = \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{1/x}$

\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)

\text{Applying L'Hôpital's Rule:}

\ln L = \lim_{x \to 0} \frac{\frac{1^x \ln 1 + 2^x \ln 2 + 3^x \ln 3 + 4^x \ln 4}{1^x + 2^x + 3^x + 4^x}}{1}

\text{Substituting } x = 0:

\ln L = \frac{1^0 \ln 1 + 2^0 \ln 2 + 3^0 \ln 3 + 4^0 \ln 4}{1^0 + 2^0 + 3^0 + 4^0}

\ln L = \frac{0 + \ln 2 + \ln 3 + \ln 4}{4}

\ln L = \frac{\ln(1 \cdot 2 \cdot 3 \cdot 4)}{4}

\ln L = \frac{\ln(4!)}{4}

\ln L = \ln(4!)^{1/4}

L = (4!)^{1/4}

$\text{Correct Option: (B)}$

Explanation

Solution:

Let $L = \lim_{x \to 0} \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{1/x}$

\ln L = \lim_{x \to 0} \frac{1}{x} \ln \left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)

\text{Applying L'Hôpital's Rule:}

\ln L = \lim_{x \to 0} \frac{\frac{1^x \ln 1 + 2^x \ln 2 + 3^x \ln 3 + 4^x \ln 4}{1^x + 2^x + 3^x + 4^x}}{1}

\text{Substituting } x = 0:

\ln L = \frac{1^0 \ln 1 + 2^0 \ln 2 + 3^0 \ln 3 + 4^0 \ln 4}{1^0 + 2^0 + 3^0 + 4^0}

\ln L = \frac{0 + \ln 2 + \ln 3 + \ln 4}{4}

\ln L = \frac{\ln(1 \cdot 2 \cdot 3 \cdot 4)}{4}

\ln L = \frac{\ln(4!)}{4}

\ln L = \ln(4!)^{1/4}

L = (4!)^{1/4}

$\text{Correct Option: (B)}$

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