Find the limit of sqrt(36x^2 + 7x + 49) - 6x as x approaches infinity.

Step-by-step solution

1. 1

   Identify the indeterminate form. As $x\to\infty$ the square root behaves like $\sqrt{36x^2}=6x$, so the expression is a difference of two quantities that both blow up:

   $\sqrt{36x^2+7x+49}\to\infty,\qquad 6x\to\infty\;\Longrightarrow\;\infty-\infty$

   That form has no value on its own; you cannot conclude $0$ from "they both grow like $6x$", because the difference of the corrections is what matters.

2. 2

   Multiply by the conjugate. The standard move for a root minus a linear term is to multiply and divide by the sum of the same two pieces:

   $\sqrt{A}-6x=\frac{\left(\sqrt A-6x\right)\left(\sqrt A+6x\right)}{\sqrt A+6x}=\frac{A-36x^2}{\sqrt A+6x},\qquad A=36x^2+7x+49$

   The numerator loses its radical entirely:

   $A-36x^2=7x+49$

3. 3

   Divide numerator and denominator by $x$. For $x>0$ we may write $\sqrt{36x^2+7x+49}=x\sqrt{36+\tfrac{7}{x}+\tfrac{49}{x^2}}$, so

   $\frac{7x+49}{\sqrt{36x^2+7x+49}+6x}=\frac{7+\frac{49}{x}}{\sqrt{36+\frac{7}{x}+\frac{49}{x^2}}+6}$

   Pulling $x$ out of the root as $+x$ (not $-x$) is valid only because $x\to+\infty$; for $x\to-\infty$ the sign would flip and the answer would change completely.

4. 4

   Take the limit of each piece. Every $\tfrac{1}{x}$ term vanishes:

   $\lim_{x\to\infty}\frac{7+\frac{49}{x}}{\sqrt{36+\frac{7}{x}+\frac{49}{x^2}}+6}=\frac{7}{\sqrt{36}+6}=\frac{7}{12}$

5. 5

   Sanity-check numerically and via the general rule. Substituting large values: at $x=10^4$ the expression is $0.583739$, at $x=10^6$ it is $0.583337$, converging on $\tfrac{7}{12}=0.58\overline{3}$ $\checkmark$. There is also a shortcut worth remembering: for $\sqrt{a^2x^2+bx+c}-ax$ with $a>0$ the limit is always $\tfrac{b}{2a}$, and here $\tfrac{7}{2\cdot 6}=\tfrac{7}{12}$.