The figure shows a right triangle with its hypotenuse $OB$ along the $y$-axis and its vertex $A$ on the parabola $y = x^2$. Let $h$ represents the length of the hypotenuse which depends on the $x$-coordinate of the point $A$. The value of $\lim_{x \to 0} (h)$ equals
Step-by-Step Solution
Key Concept: Identify the floor function value by establishing bounds, then rationalize expressions with square roots by dividing by the leading term.
Since $n < \sqrt{n^2+n+1} < n+1$, we have $[\sqrt{n^2+n+1}] = n$. To find the limit, rationalize: $\lim_{n\to\infty}\frac{h+1}{\sqrt{n^2+n+1}+n} = \lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n+1}+n}$. Dividing numerator and denominator by $n$, this becomes $\frac{1+1/n}{\sqrt{1+1/n+1/n^2}+1}\to\frac{1}{2}$.
Correct Answer: 2