Limit of $\frac{1}{M} \int_1^M M^{\frac{1}{x}} \textrm{d} x$ when $M \to +\infty$

MacLaurin Expansion of $m^y$ is $$m^y=1+y \log m+O\left(y^2\right)$$ Therefore $$m^{1/x}=1+\frac{\log m}{x}+O\left(1/x^2\right)\tag{1}$$ Integrate $$\int_1^m \left(\frac{\log m}{x}+1\right) \, dx=m+\log ^2 m-1\tag{2}$$ Thus $$\underset{m\to \infty }{\text{lim}}\frac{\int_1^m m^{1/x} \, dx}{m}=\underset{m\to \infty }{\text{lim}}\frac{m+\log ^2 m-1}{m}=\underset{m\to \infty }{\text{lim}}\left(1+\frac{\log ^2 m}{m}-\frac{1}{m}\right)=1$$

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Edit

$(1)$

$$\int_1^m \frac{1}{x^2} \, dx=1-\frac{1}{m}$$ $(2)$ becomes $$\int_1^m \left[\left(\frac{\log m}{x}+1\right) +O\left(1/x^2\right)\right]\, dx\le m+\log ^2 m-1 +1-\frac{1}{m}=m+\log ^2 m-\frac{1}{m}$$ Thus to a greater extent $(2)$ holds