Limit of this expression for apparent length

[etotheipi]

Homework Statement

See below

Relevant Equations

N/A

The apparent length of a rod is determined to be$$\tilde{L}(x_0) = \gamma L + \beta \gamma \sqrt{D^2 + (\gamma x_0 - \frac{L}{2})^2} - \beta \gamma \sqrt{D^2 + (\gamma x_0 + \frac{L}{2})^2}$$I am trying to determine expressions for $\tilde{L}(x_0)$ when $x_0 \rightarrow -\infty$ and $x_0 \rightarrow \infty$. I rearranged it to$$\tilde{L} = \gamma L + \beta \gamma (\gamma x_0 - \frac{L}{2})\sqrt{1+\frac{D^2}{(\gamma x_0 -\frac{L}{2})^2}} - \beta \gamma (\gamma x_0 + \frac{L}{2})\sqrt{1+\frac{D^2}{(\gamma x_0 +\frac{L}{2})^2}}$$And it would then seem in both cases, since the square roots both approach 1 in both limits, that$$\tilde{L}(x_0 \rightarrow -\infty) = \tilde{L}(x_0 \rightarrow \infty)= \gamma L(1-\beta)$$However it is noted in the solution manual that in fact $$\tilde{L}(x_0 \rightarrow -\infty) =\gamma L(1+\beta)$$ and $$\tilde{L}(x_0 \rightarrow \infty) =\gamma L(1-\beta)$$What have I missed? Thanks!

 

[Infrared]

Careful with signs: $\sqrt{1+a^2}=|a|\sqrt{(1/a^2)+1}.$

 

[etotheipi]

Careful with signs: $\sqrt{1+a^2}=|a|\sqrt{(1/a^2)+1}.$

Ah, yes that's a little subtle... now I'm going to be quite paranoid about this in the future . It should then be $$\tilde{L} \approx \gamma L + \beta \gamma \left( \left|\gamma x_0 - \frac{L}{2} \right| - \left|\gamma x_0 + \frac{L}{2} \right| \right)$$and that does give the correct result when evaluated. Thanks!