Exercise 8.4

Question

Exercise 4: 
    Prove the following limit relations:
    \begin{itemize}
        \item[(a)] $\displaystyle\lim_{x\rightarrow0}\frac{b^x-1}{x}=\log(b)\quad(b>0)$.
        \item[(b)] $\displaystyle\lim_{x\rightarrow0}\frac{\log(1+x)}{x}=1$.
        \item[(c)] $\displaystyle\lim_{x\rightarrow0}(1+x)^{1/x}=e$.
        \item[(d)] $\displaystyle\lim_{n\rightarrow\infty}\bigg(1+\frac{x}{n}\bigg)^n=e^x$.
    \end{itemize}

analambanomenos · Accepted Answer

(a) By L'Hospital's Rule, Theorem 5.13, $$\lim_{x\rightarrow 0}\frac{b^x-1}{x}=\lim_{x\rightarrow 0}\frac{(\log b)b^x}{1}=\log b.$$

(b) By L'Hospital's Rule, $$\lim_{x\rightarrow 0}\frac{\log(1+x)}{x}=\lim_{x\rightarrow 0}\frac{1/(1+x)}{1}=1.$$

Since the exponential function is continuous, if $g(x)$ is a continuous function, and $f(x)=\log\big(g(x)\big)$, and $\lim_{x\rightarrow a}f(x)=A$, then $\lim_{x\rightarrow a}g(x)=e^A$.

(c) Let $f(x)=\log(1+x)^{1/x}=\big(\log(1+x)\big)/x$. Then, from part (b), $$\displaystyle\lim_{x\rightarrow0}(1+x)^{1/x}=e^{\lim_{x\rightarrow 0}f(x)}=e^1=e.$$

(d) Let $f(x)=\log\big(1+(x/n)\big)^n=n\log\big(1+(x/n)\big)$. By L'Hospital's Rule $$\lim_{n\rightarrow\infty}f(x)=\lim_{n\rightarrow\infty}\frac{\log\bigg(1+\displaystyle
    \frac{x}{n}\bigg)}{1/n}=\lim_{n\rightarrow\infty}\frac{-\displaystyle\frac{x}{n^2}\bigg(\frac{1}{1+x/n}\bigg)}{-1/n^2}=\lim_{n\rightarrow\infty}\frac{x}{1+x/n}=x.$$ Hence
    $\lim_{n\rightarrow\infty}\big(1+(x/n)\big)^n=e^x$.

Exercise 8.4

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Exercise 8.4

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