Find the limit as x approaches infinity of sqrt((36x^2 - 25x + 16)/(9x^2 - 4x + 1)).

Step-by-step solution

1. 1

   Handle the inside first. The square root is continuous on $[0,\infty)$, so

   $\lim_{x\to\infty}\sqrt{g(x)} = \sqrt{\lim_{x\to\infty} g(x)}$

   provided the inner limit exists and is non-negative. So evaluate the rational function, then take the root once at the end.

2. 2

   Divide top and bottom by the highest power, $x^2$. This is the standard move for a limit at infinity, and it converts every lower-order term into something that vanishes:

   $\frac{36x^2-25x+16}{9x^2-4x+1} = \frac{36 - \dfrac{25}{x} + \dfrac{16}{x^2}}{9 - \dfrac{4}{x} + \dfrac{1}{x^2}}$

3. 3

   Let the reciprocal terms go to zero. As $x\to\infty$, each of $\tfrac{25}{x}$, $\tfrac{16}{x^2}$, $\tfrac{4}{x}$ and $\tfrac{1}{x^2}$ tends to $0$:

   $\lim_{x\to\infty}\frac{36x^2-25x+16}{9x^2-4x+1} = \frac{36}{9} = 4$

   Since numerator and denominator have equal degree, the limit is just the ratio of leading coefficients.

4. 4

   Take the square root.

   $\sqrt{4} = 2$

   The answer is $+2$, the principal root; a $\pm$ would be wrong because the expression under discussion is a non-negative square root throughout.

5. 5

   Guard against the tempting shortcut. Writing $\sqrt{36x^2}/\sqrt{9x^2} = 6x/3x = 2$ happens to give the right value here, but only because both leading coefficients are perfect squares and $x > 0$. The safe route is always to reduce the inside first.

6. 6

   Verify numerically. At $x = 1000$: numerator $= 36{,}000{,}000 - 25{,}000 + 16 = 35{,}975{,}016$, denominator $= 9{,}000{,}000 - 4{,}000 + 1 = 8{,}996{,}001$, ratio $= 3.999003$, square root $= 1.999751$. At $x = 10^6$ the root is $1.99999975$ ✓ — converging to $2$, and a symbolic limit returns exactly $2$.