Juan Hewlett

2021-12-30

Proving an inequality without an integral:

$\frac{1}{x+1}\le \mathrm{ln}(1+x)-\mathrm{ln}\left(x\right)\le \frac{1}{x}$

ol3i4c5s4hr

Beginner2021-12-31Added 48 answers

Let $f\left(x\right){\textstyle \phantom{\rule{0.222em}{0ex}}}=\mathrm{ln}\left(x\right),$ , then the Mean Value Theorem (Differentiation) says that there exists some $\xi \in (x,x+1)$

$\mathrm{ln}(1+x)-\mathrm{ln}\left(x\right)=\frac{\mathrm{ln}(1+x)-\mathrm{ln}\left(x\right)}{(1+x)-x}=\frac{f(1+x)-f\left(x\right)}{(1+x)-x}=\frac{df}{dx}\xi$

As$\frac{df}{dx}\left(x\right)=\frac{1}{x}$ and as the function $\frac{1}{x}$ is monotonic we know that $\frac{1}{x+1}\le \frac{df}{dx}\left(\xi \right)\le \frac{1}{x}$

As

jean2098

Beginner2022-01-01Added 38 answers

If we write $t=\frac{1}{x}$ the inequalities become

$\frac{t}{t+1}\le \mathrm{ln}(1+t)\le t,t>0$

Observe that the three expressions are equal when$t=0$ , and then show, by differentiating, that $t-\mathrm{ln}(1+t)$ and $\mathrm{ln}(1+t)-\frac{t}{1+t}$ are increasing functions.

Observe that the three expressions are equal when

Vasquez

Expert2022-01-09Added 669 answers

This is trivial using the definition of

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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A) No

B) 0

C) Yes

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A)$1.496\times {10}^{11}$

B)$1.496\times {10}^{10}$

C)$1.496\times {10}^{12}$

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