# Two limits with hyperbolic tangent

I'm not very experienced with double limits and now I have to evaluate the limits $$\lim_{x \to \infty,\ y \to \infty} \left[ \tanh(x+2y) + \tanh(x-2y) \right]$$ $$\lim_{x \to \infty,\ y \to \infty} \left[ \tanh(x+2y) - \tanh(x-2y) \right]$$ and I'm really not sure how to proceed with this. Naturally the first terms are $$\lim_{x \to \infty,\ y \to \infty} \tanh(x+2y) = 1$$ but that's as far as I have gotten.

Does this limit exist?

EDIT: So after a little work I can show that $$\tanh(x-2y) = \frac{\sinh(x-2y)}{\cosh(x-2y)} = \frac{e^{x-2y} - e^{-(x-2y)}}{e^{x-2y} + e^{-(x-2y)}} = \frac{e^{-2y} - e^{-2x}e^{2y}}{e^{-2y} + e^{-2x} e^{2y}}$$ which for constant $y < \infty$ behaves as $$\lim_{x \to \infty} \tanh(x-2y) = \frac{e^{-2y} - e^{-2x}e^{2y}}{e^{-2y} + e^{-2x} e^{2y}} = \frac{e^{-2y} - 0 \cdot e^{2y}}{e^{-2y} + 0 \cdot e^{-y}} = 1$$ but also $$\lim_{y \to \infty} \tanh(x-2y) = \frac{e^{x-4y} - e^{-x}}{e^{x-4y} + e^{-x}} = \frac{0 - e^{-x}}{0 + e^{-x}} = - 1 \mathrm{.}$$

However as $y \to \infty$ AND $x \to \infty$ there's the term $e^{-2x}e^{2y}$ where the two terms tend to zero and infinity. What is the specific reason that allows me to say that this limit doesn't exist?

• Is it $+\infty$ or $-\infty$ – hamam_Abdallah May 26 '17 at 15:13
• It is $+ \infty$ – 655321 May 26 '17 at 15:15

\begin{align}\tanh(x+2y)+\tanh(x-2y)&=\frac{\sinh(x+2y)}{\cosh(x+2y)}+\frac{\sinh(x-2y)}{\cosh(x-2y)}\\&=\frac{\sinh(x+2y)\cosh(x-2y)+\sinh(x-2y)\cosh(x+2y)}{\cosh(x+2y)\cosh(x-2y)}\\&=\frac{\sinh(2x)}{\cosh(x+2y)\cosh(x-2y)}\\&=\frac{\frac{e^{2x}-e^{-2x}}2}{\frac{(e^{x+2y}+e^{-(x+2y)})(e^{x-2y}+e^{-(x-2y)})}4}\\&=2\times\frac{1-e^{-6x}}{(e^{2y}+e^{-2x-2y})(e^{-2y}+e^{-2x-2y})}\end{align}
Then note that as $x\to\infty$ and $y$ is held constant, then it goes to $2$. Likewise, as $y\to\infty$ and $x$ is held constant, then it goes to $0$. Thus, the limit doesn't exist.
• But if $y$ is held constant ($< \infty$) isn't $\lim_{x \to \infty} \left[ \tanh(x+2y) + \tanh(x-2y) \right] = 1 + 1 = 2$? And not infinity? – 655321 May 26 '17 at 15:28
• Yes but $\sinh(x+2y) \cosh(x-2y) + \sinh(x-2y) \cosh(x+2y) = \sinh(2x) \neq \sinh(4x)$ and the last term in the denominator should be $e^{-(2x-2y)} = e^{-2x+2y}$ so $$\tanh(x+2y) - \tanh(x-2y) = 2 \cdot \frac{1-e^{-4x}}{(e^{2y} + e^{-2x-2y})(e^{-2y}+e^{-2x+2y})} \mathrm{.}$$ I find it hard to believe that a sum on $\tanh$-functions would tend to infinity as $\tanh(x) \in (-1, 1)$. Anyway I think that this limit really doesn't exist. – 655321 May 26 '17 at 17:08