Application of Derivatives... Problem... Prove that ln (1+x) 0 .

Question

Application of Derivatives... 
Problem...

Prove that ln (1+x) < x for x > 0 .

mathslover · Answer

@BSwan and @ganeshie8

anonymous · Answer

so u need to prove that
ln (1+x) < x 
x - ln (1+x)>=0
 or
ln (1+x)-x<0

mathslover · Answer

Yep, I think we need to prove 

$\ln(1+x) - x < 0$ or $-\ln (1+x) + x > 0$

mathslover · Answer

Okay, got it.

anonymous · Answer

yep , ok how did u got it ?

mathslover · Answer

f(X) = ln(1+x) - x

f'(x) = $\cfrac{1}{1+x} - 1$ 

$\cfrac{1 - (1+x)}{1+x}$ 

$\cfrac{-x}{1+x}$ 

now as f'(x) > 0 (for x >0) 

therefore,

$\cfrac{-x}{1+x} < 0$  (for x > 0)

mathslover · Answer

f(x) is thus, decreasing for x > 0.

f(x) < f(0) 

$\ln(1+x) - x < f(0)$ 

$\ln (1+x) < x$

mathslover · Answer

This is how I did it...

anonymous · Answer

well mmm
but it not tells u it dnt cross the x axis :O
|dw:1401815426099:dw|

mathslover · Answer

Why does it need to cross the x-axis? (sorry, I am yet to learn the geometrical representations in this chapter)

anonymous · Answer

well if its >0 then it dnt cross the x axis

mathslover · Answer

Okay yeah as x > 0 then it should not cross the x axis...

mathslover · Answer

Is that graph for ln(1+x) < x (for all x > 0) ?

anonymous · Answer

@mathslover, your method is right. I would have done it slightly differently.

Let $f(x)=\ln(1+x)$ and $g(x)=x$. We know that $f(0)=g(0)=0$, so to show that $f(x)<g(x)$ we need to show that $g(x)$ is increasing faster than $f(x)$. To do so, we must show $f'(x)<g'(x)$:
$$\begin{cases}f'(x)=\dfrac{1}{1+x}\\g'(x)=1\end{cases}$$
Clearly, for all $x>0$, we have $\dfrac{1}{1+x}<1$ since $\dfrac{1}{1+x}\to0$ as $x\to\infty$. And we're done.

mathslover · Answer

@SithsAndGiggles -> That was a great method. I appreciate that! Thanks a lot.