Prove that xx+1&lt;ln⁡(1+x)&lt;x for x&gt;−1,x≠0

Let f(x)=log⁡(1+x) for x&gt;−1. Then f′(x)=11+x and the mean-value theorem states that for x≠0 f(x)=f(0)+(x−0)f′(c)=x1+c for some c strictly between 0 and x. If x&gt;0 then 0&lt;c&lt;x⇒11+x&lt;11+c&lt;1, and if x&lt;0 then x&lt;c&lt;0⇒11+x&gt;11+c&gt;1. In both cases, multiplying the inequalities by x gives x1+x&lt;f(x)=x1+c&lt;x.

Let f(x)=x−ln⁡(1+x), where x&gt;−1. f′(x)=1−11+x=x1+x, which says that f(x)≥f(0)=0. In another hand, let g(x)=ln⁡(1+x)−xx+1, where x&gt;−1. Thus, g′(x)=11+x−1(1+x)2=x(1+x)2, which gives again that xmin=0 and we obtain g(x)≥g(0)=0. Since, x≠0 we get xx+1&lt;ln⁡(1+x)&lt;x. Done!

Let f(x)=log⁡(1+x) for x&gt;−1. Then f′(x)=11+x and the mean-value theorem states that for x≠0 f(x)=f(0)+(x−0)f′(c)=x1+c for some c strictly between 0 and x. If x&gt;0 then 0&lt;c&lt;x⇒11+x&lt;11+c&lt;1, and if x&lt;0 then x&lt;c&lt;0⇒11+x&gt;11+c&gt;1. In both cases, multiplying the inequalities by x gives x1+x&lt;f(x)=x1+c&lt;x.

Let f(x)=x−ln⁡(1+x), where x&gt;−1. f′(x)=1−11+x=x1+x, which says that f(x)≥f(0)=0. In another hand, let g(x)=ln⁡(1+x)−xx+1, where x&gt;−1. Thus, g′(x)=11+x−1(1+x)2=x(1+x)2, which gives again that xmin=0 and we obtain g(x)≥g(0)=0. Since, x≠0 we get xx+1&lt;ln⁡(1+x)&lt;x. Done!

"Prove that xx+1<ln⁡(1+x)<x for x>−1,x≠0" was answered by our experts

William Boggs

Answered question

2022-01-20

Prove that

\frac{x}{x + 1} < \ln (1 + x) < x

for

x > - 1, x \neq 0

Answer & Explanation

Becky Harrison

Beginner2022-01-20Added 40 answers

Let

f (x) = \log (1 + x)

for

x > - 1

. Then

f' (x) = \frac{1}{1 + x}

and the mean-value theorem states that for

x \neq 0

f (x) = f (0) + (x - 0) f' (c) = \frac{x}{1 + c}

for some

c

strictly between

0

and

x

.
If

x > 0

then

0 < c < x \Rightarrow \frac{1}{1 + x} < \frac{1}{1 + c} < 1,

and if

x < 0

then

x < c < 0 \Rightarrow \frac{1}{1 + x} > \frac{1}{1 + c} > 1 .

In both cases, multiplying the inequalities by

x

gives

\frac{x}{1 + x} < f (x) = \frac{x}{1 + c} < x .

porschomcl

Beginner2022-01-21Added 28 answers

Let

f (x) = x - \ln (1 + x)

, where

x > - 1

f' (x) = 1 - \frac{1}{1 + x} = \frac{x}{1 + x}

, which says that

f (x) \geq f (0) = 0

.
In another hand, let

g (x) = \ln (1 + x) - \frac{x}{x + 1}

, where

x > - 1

.
Thus,

g' (x) = \frac{1}{1 + x} - \frac{1}{{(1 + x)}^{2}} = \frac{x}{{(1 + x)}^{2}}

, which gives again
that

x_{min} = 0

and we obtain

g (x) \geq g (0) = 0 .

Since,

x \neq 0

we get

\frac{x}{x + 1} < \ln (1 + x) < x .

Done!

RizerMix

Expert2022-01-27Added 656 answers

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