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[SOLVED] Limit of a vector of two variables

dwsmith

Well-known member
Feb 1, 2012
1,673
\begin{align*}
\mathbf{u}_1 &= h_u\mathbf{U}_u\\
&= \frac{a\left(\sinh(u)\cos(v)\unit{i} +
\cosh(u)\sin(v)\unit{j}\right)}{a\sqrt{\cosh^2(u) - \cos^2(v)}}\\
&= \frac{\sinh(u)\cos(v)\unit{i} + \cosh(u)\sin(v)\unit{j}}
{\sqrt{\cosh^2(u) - \cos^2(v)}}\\
\mathbf{u}_2 &= h_v\mathbf{U}_v\\
&= \frac{-a\left(\cosh(u)\sin(v)\unit{i} +
\sinh(u)\cos(v)\unit{j}\right)}{a\sqrt{\cosh^2(u) - \cos^2(v)}}\\
&= \frac{-\cosh(u)\sin(v)\unit{i} + \sinh(u)\cos(v)\unit{j}}
{\sqrt{\cosh^2(u) - \cos^2(v)}}
\end{align*}

I want to take the limit as \(u\to\infty\) and show that \(\mathbf{u}_1\sim\cos(v)\unit{i} + \sin(v)\unit{j}\) and \(\mathbf{u}_2\sim -\sin(v)\unit{i} + \cos(v)\unit{j}\).

I did this but I don't like just neglecting the v terms and then just putting it back in as if nothing happened.

\begin{align*}
\lim\limits_{u\to\infty}\frac{\sinh(u)}{\sqrt{\cosh^2(u) -
\cos^2(v)}} &=
\lim\limits_{u\to\infty}\frac{\sinh(u)}{\cosh(u)}\\
&= \lim\limits_{u\to\infty}\frac{e^u - e^{-u}}{e^u + e^{-u}}\\
&= \lim\limits_{u\to\infty}\frac{e^{2u} - 1}{e^{2u} + 1}\\
&= \lim\limits_{u\to\infty}\frac{e^{2u}}{e^{2u}}\\
&= 1\\
\lim\limits_{u\to\infty}\frac{\cosh(u)}{\sqrt{\cosh^2(u) -
\cos^2(v)}} &=
\lim\limits_{u\to\infty}\frac{\cosh(u)}{\cosh(u)}\\
&= 1
\end{align*}