Calculus problem #2130

We notice that since we want limit as $x\rightarrow \infty$ , we can only consider integer values of $x$ . Then we get:

$\lfloor x^{2}+x\rfloor = x^{2}+x$ .

After this simplification, we can write,

$N=Lim_{x\rightarrow \infty}x(1+\frac{1}{x})^{\frac{1}{2}}-x$

$\therefore N=Lim_{x\rightarrow \infty}\frac{(1+\frac{1}{x})^{\frac{1}{2}}-1}{\frac{1}{x}}$

Now using L'Hospital's Rule , we get,

$N=Lim_{x\rightarrow \infty}\frac{\frac{1}{2}(1+\frac{1}{x})^{\frac{-1}{2}}(\frac{-1}{x^{2}})}{\frac{-1}{x^{2}}}$

This gives, $N=\boxed{\frac{1}{2}}$

We open up the floor function such that $\left \lfloor x^{2}+x \right \rfloor=x^{2}+x+k$ , where $k$ is the decimal part of $x^2+x$ . Then: $\lim_{x\rightarrow \infty }(\sqrt{\left \lfloor x^{2}+x \right \rfloor}-x)=\lim_{x\rightarrow \infty }(\sqrt{x^{2}+x+k}-x)$ Multiplying with its conjugate, we get: $\lim_{x\rightarrow \infty }(\sqrt{x^{2}+x+k}-x)=\lim_{x\rightarrow \infty }(\frac{{x^{2}+x+k}-x^{2}}{\sqrt{x^{2}+x+k}+x})$ Then, we can circumvent this equation by dividing all terms with the highest power of $x$ in the denominator. With that, we obtain: $\lim_{x\rightarrow \infty }(\frac{{x^{2}+x+k}-x^{2}}{\sqrt{x^{2}+x+k}+x})=\lim_{x\rightarrow \infty }(\frac{\frac{1}{x}(x+k)}{\frac{1}{x}(\sqrt{x^{2}+x+k}+x)})$ $=\lim_{x\rightarrow \infty }(\frac{1+\frac{k}{x}}{\sqrt{1+\frac{1}{x}+\frac{k}{x^{2}}}+1})$ Since $x\rightarrow \infty$ , then $\frac{1}{x}\rightarrow 0$ , $\therefore \lim_{x\rightarrow \infty }(\frac{1+\frac{k}{x}}{\sqrt{1+\frac{1}{x}+\frac{k}{x^{2}}}+1})=\frac{1}{\sqrt{1}+1}$ Our desired answer is $\boxed{\frac{1}{2}}$ .

Substitute 99999 for infinity, then the expression will yield 0.499999...

The question states that the given limit exists over the set of real numbers. Thus, it implies that the limit would also exist if we constraint the variable x to move only through the set of natural numbers as it reaches infinity. So the problem can be reworded to N= \limit\n\infinity (sqrt(n^2+n)-n). This on rationalization reduces to n/sqrt(n^2+n) + n. This yields 0.5