Limit of a difference of square roots over sin(2x)
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{\sqrt{1-\tan x} - \sqrt{1+\tan x}}{\sin(2x)}. $$
Explanation of Method
Direct substitution gives $\frac{1-1}{0}=\frac00$. Rationalizing the numerator by multiplying by its conjugate collapses the difference of two square roots into a difference of squares, $(1-\tan x)-(1+\tan x)=-2\tan x$, which is a much simpler expression. Combined with the small-angle approximations $\tan x\approx x$ and $\sin(2x)\approx2x$, the whole limit reduces to a ratio of simple linear terms in $x$.
- Rationalize the numerator by multiplying by its conjugate.
- Simplify using the difference of squares.
- Approximate each remaining piece using $\tan x\approx x$ and $\sin(2x)\approx2x$ for small $x$.
- Combine and take the limit.
Step 1 · Rationalize the numerator
Multiply the numerator (and, to keep the expression unchanged, an equivalent factor in the denominator) by the conjugate $\sqrt{1-\tan x}+\sqrt{1+\tan x}$:
$$ \sqrt{1-\tan x}-\sqrt{1+\tan x} = \frac{\left(\sqrt{1-\tan x}-\sqrt{1+\tan x}\right)\left(\sqrt{1-\tan x}+\sqrt{1+\tan x}\right)}{\sqrt{1-\tan x}+\sqrt{1+\tan x}}. $$
Step 2 · Apply the difference of squares
The numerator of the right-hand side becomes:
$$ \left(\sqrt{1-\tan x}\right)^2 - \left(\sqrt{1+\tan x}\right)^2 = (1-\tan x)-(1+\tan x) = -2\tan x. $$
So:
$$ \sqrt{1-\tan x}-\sqrt{1+\tan x} = \frac{-2\tan x}{\sqrt{1-\tan x}+\sqrt{1+\tan x}}. $$
Step 3 · Apply small-angle approximations
As $x\to0$, $\tan x\to0$, so both $\sqrt{1-\tan x}\to1$ and $\sqrt{1+\tan x}\to1$, giving:
$$ \sqrt{1-\tan x}+\sqrt{1+\tan x} \to 1+1 = 2. $$
Also, $\tan x \approx x$ for small $x$. Substituting both approximations:
$$ \sqrt{1-\tan x}-\sqrt{1+\tan x} \approx \frac{-2x}{2} = -x. $$
Step 4 · Combine and take the limit
Using $\sin(2x)\approx2x$ for small $x$, the full limit becomes:
$$ L = \lim_{x\to0}\frac{-x}{2x} = -\frac{1}{2}. $$
Summary Table
| Quantity | Simplified / approximated form | Behavior as $x \to 0$ |
|---|---|---|
| Numerator (rationalized) | $\dfrac{-2\tan x}{\sqrt{1-\tan x}+\sqrt{1+\tan x}}$ | $\approx -x$ |
| Denominator | $\sin(2x)$ | $\approx 2x$ |
| Ratio | — | $\to -\dfrac12$ |
Additional Notes
- Rationalizing first is what converts a difference of two square roots — awkward to approximate directly — into the single, simple expression $-2\tan x$ divided by a sum that safely tends to a nonzero constant, $2$.
- This limit can also be checked with L'Hôpital's Rule directly on the original expression, differentiating $\sqrt{1-\tan x}-\sqrt{1+\tan x}$ and $\sin(2x)$ and substituting $x=0$, which also yields $-\frac12$ — the rationalization method here avoids that extra differentiation step.
- The technique generalizes: any limit of the form $\lim_{x\to0}\dfrac{\sqrt{1-g(x)}-\sqrt{1+g(x)}}{h(x)}$ with $g(x)\to0$ reduces, after rationalizing, to $\dfrac{-g(x)}{h(x)}$ in the limit, since the conjugate sum always tends to $2$.
Result
The value of the limit is: