Why is $\lim_{x \to \infty}(\int_0^n k^{1-x}\,\,di)^{1/(1-x)} = k$ when $k,n>0$ is constant real? Why is $$\lim_{x \to \infty}\left(\int_0^n k^{1-x}\,\,di\right)^{1/(1-x)} = k$$
?
$k,n>0$ is constant real.
 A: $$Q = \int_0^n k^{1-x}\ di = \left[ik^{1-x}\right]_0^n = nk^{1-x}$$
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$$\text{Now, consider the following:}$$
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$$\lim_{x \to \infty} Q^{\large\frac{1}{1-x}} = \lim_{x\to\infty}\exp \left(\dfrac{\ln\left(nk^{1-x}\right)}{1-x}\right) = \lim_{x\to\infty}\exp\bigg(\dfrac{\ln(n)}{1-x}\bigg) \cdot \exp\bigg(\ln(k) \bigg)  = k$$
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A: Here are the steps
$$\lim_{x \to \infty}\left[\int_0^n k^{1-x}\ di\right]^{\frac{1}{1-x}}= \lim_{x \to \infty}\left[ k^{1-x}\int_0^n di\right]^{\frac{1}{1-x}} $$
$$ = \lim_{x \to \infty}\left[ k^{1-x}(n-0)\right]^{\frac{1}{1-x}}= \lim_{x \to \infty}\left[ nk^{1-x}\right]^{\frac{1}{1-x}} $$
$$ = \lim_{x \to \infty}e^{\ln\left(nk^{1-x}\right)^{\frac{1}{1-x}}}= \lim_{x \to \infty}e^{\frac{\ln\left(nk^{1-x}\right)}{1-x}} $$
$$ = \lim_{x \to \infty}e^{\frac{\ln(n)+\ln\left(k^{1-x}\right)}{1-x}} = e^{\lim\limits_{x\to\infty}\frac{\ln(n)+(1-x)\ln(k)}{1-x}} $$
$$ = e^{\lim\limits_{x\to\infty}\left[\frac{\ln(n)}{1-x}+\ln(k)\right]}= e^{0+\ln(k)} =k$$
A: You only need to compute the integral and use the laws of exponents. Note that
$$
\int _0^n k^{1-x}di= nk^{1-x}
$$
And use the fact that $\lim\limits_{x\to \infty }\frac{1}{1-x}=0$.
