How is this proof that [MATH]\sqrt(1+x^2)-x>0.[/MATH]?

[333happy]

Let [MATH]f(x)=\sqrt{1+x^2}-x.[/MATH] One can easily see that [MATH]f(x)\neq 0[/MATH], so either [MATH]f(x)>0[/MATH] or [MATH]f(x)<0[/MATH]. Since [MATH]\lim_{x\to -\infty}f(x)=\infty[/MATH] and [MATH]\lim_{x\to \infty}f(x)=0[/MATH], [MATH]f(x)\in(-\infty,0)[/MATH], whence [MATH]f(x)>0.[/MATH]

 

[Steven G]

What you said is not necessary true if f(x) is not continuous. So to use your method you must show that f(x) is continuous. Why with all these limits? You could have just used the 1st limit or you could have just computed f(1) = sqrt(2) - 1 >0 and be done.

 

[Steven G]

Just for the record, if f(x) is in (-oo, 0) then f(x) <0!!!!

 

[Steven G]

I was also taught that if something is easily seen then you should prove it! Maybe not in your case, but usually when a student writes that something is easily seen it means that they do not know how to prove it.

 

[JeffM]

One incredibly easy way to prove it is this.

[MATH]\text {Given } f(x) = \sqrt{x^2 + 1} - x \text { and } x \in \mathbb R.[/MATH]
[MATH]\text {Assume, for purposes of contradiction, that }\\ \exists \text { real } r \ | \ \sqrt{r^2 + 1} \le r.[/MATH][MATH]\text {By definition, } 0 \le \sqrt{r^2 + 1} \implies\\ 0 \le \sqrt{r^2 + 1} * \sqrt{r^2 + 1} \le \sqrt{r^2 + 1} * r = r\sqrt{r^2 + 1 } \le r^2 \implies \\ r^2 + 1 \le r^2 \implies 1 \le 0, \text { which is absurd} \implies \\ \not \exists \text { real } r \ | \ \sqrt{r^2 + 1} \le r \implies \text{WHAT?}[/MATH]