Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) I had this problem in an exam I recently appeared for:

Find the range of
$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$

By randomly assuming the value of $x$, I got the lower range of this expression as $3/7$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.
Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.
Another guy I met outside the exam center, told me he used an approach of $x$ tending to infinity in both cases and got the maximum value of this expression as $1/2$. But before I could ask him to explain more on this method, he had to leave for his work.
So, will someone please throw some light on this method of $x$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).

In general, if $\deg f = 0$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $f$ as $x$ increases/decreases without bound is $a_n/b_n$.
In your case, $a_2 = 1$ and $b_2 = 2$. Hence, $a_2/b_2 = 1/2$.

We'll factor $f$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$
Notice that for all $x \in \mathbb{R}$, $f > 0$. Also, we can see that $(x+1)^2 + 2 < 2(x + 1)^2 + 7$. This means that the range should be a part of $(0,1/2)$. Since both numerator and denominator have $(x + 1)^2$ without any remaining $x$'s, we can see that this will be at its minimum when $x = -1$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$
Therefore, the range is $[3/7, 1/2)$.
