Aspire Faculty ID #11624 · Topic: NIMCET 2024 · Just now
NIMCET 2024

The value of the limit $$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$$ is

Solution

We need to find: $\displaystyle \lim_{x\to 0}\left( \frac{1^x + 2^x + 3^x + 4^x}{4} \right)^{1/x}$ 
Rewrite each term: $k^x = e^{x\ln k}$ 
So: $1^x + 2^x + 3^x + 4^x = e^{0} + e^{x\ln 2} + e^{x\ln 3} + e^{x\ln 4}$ 
As $x \to 0$ use expansion: $e^{x\ln k} = 1 + x\ln k + O(x^2)$ 
So: $1^x + 2^x + 3^x + 4^x$ 
$= 1 + (1 + x\ln 2) + (1 + x\ln 3) + (1 + x\ln 4)$ $= 4 + x(\ln 2 + \ln 3 + \ln 4) + O(x^2)$ 
Thus: $\frac{1^x + 2^x + 3^x + 4^x}{4} = 1 + \frac{x(\ln 2 + \ln 3 + \ln 4)}{4} + O(x^2)$ 
Now apply the standard limit: 
$\lim_{x\to 0} (1 + kx)^{1/x} = e^{k}$ 
Here: $k = \frac{\ln 2 + \ln 3 + \ln 4}{4}$ 
Hence the limit is: $e^{(\ln 2 + \ln 3 + \ln 4)/4}$ 
Combine logs: $\ln 2 + \ln 3 + \ln 4 = \ln(2\cdot 3 \cdot 4) = \ln 24$ 
Final answer: $\boxed{24^{1/4}}$

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