Method of Rationalization

If the expression inside a limit contains a square root (or, later, a trigonometric function), it can often be simplified by multiplying numerator and denominator by the radical's rationalizing (conjugate) factor — this clears the square root from whichever side is producing the 0/0 form, converting it into an ordinary polynomial ratio that can be handled by cancellation.\n\nWorked pattern 1 (single radical): $\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x}$: multiplying top and bottom by $\sqrt{1+x}+1$ turns the numerator into $(1+x)-1=x$, which cancels the x in the denominator, leaving $\lim_{x\to0}\dfrac{1}{\sqrt{1+x}+1}=\dfrac12$.\n\nWorked pattern 2 (a radical on both sides of a difference): $\lim_{z\to0}\dfrac{(b+z)^{1/2}-(b-z)^{1/2}}{z}$: multiplying by the conjugate $\sqrt{b+z}+\sqrt{b-z}$ collapses the numerator to $(b+z)-(b-z)=2z$, so the z's cancel and the limit is $\dfrac{2}{\sqrt{b+0}+\sqrt{b-0}}=\dfrac{2}{2\sqrt b}=\dfrac{1}{\sqrt b}$.\n\nWorked pattern 3 (two different radicands needing the SAME conjugate trick, then a second factoring pass): $\lim_{x\to4}\dfrac{x^2+x-20}{\sqrt{x^2-7}-\sqrt{25-x^2}}$: multiplying by the conjugate $\sqrt{x^2-7}+\sqrt{25-x^2}$ turns the denominator into $(x^2-7)-(25-x^2)=2x^2-32=2(x-4)(x+4)$; the numerator factors as $(x-4)(x+5)$; the shared $(x-4)$ cancels, leaving $\dfrac{(x+5)(\sqrt{x^2-7}+\sqrt{25-x^2})}{2(x+4)}$, which evaluates at $x=4$ to $\dfrac{(9)(\sqrt9+\sqrt9)}{2(8)}=\dfrac{9\times6}{16}=\dfrac{27}{8}$.\n\nThe same rationalizing idea extends to expressions where the variable itself sits under a square root in the denominator (as in $\lim_{x\to1}\dfra …