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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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