Let $u(x,t)$ be the bounded solution of $\frac{\partial u}{\partial t} -\frac{\partial ^2 u}{\partial x^2}=0$ with $u(x, 0)=\dfrac{e^ {2x} -1}{e^{2x} +1}.$ Then $\lim _{t \rightarrow \infty}u(1, t)=$
$$(A)-\frac{1}{2}~~~~~~(B)\frac{1}{2}~~~~~~(C)-1~~~~~~(D)1.$$
By separation of variable the solution of the above PDE is: $$u(x, t)=\begin{cases} e^{- \lambda ^2 t}(c_1 \cos(\lambda x)+c_2 \sin(\lambda x)) & ~~~\text{if}~k=-{\lambda}^2<0 ,\\ e^{\lambda ^2 t}(c_1 e^{\lambda x}+c_2 e^{\lambda x}) & ~~~\text{if}~k={\lambda}^2>0 ,\\ c_1x+c_2 & ~~~\text{if}~k=0 .\\ \end{cases}$$ where $k$ is separation constant. But when $t=0$, I cannot compare the solution with given $u(x,0)$, that's why I cannot find $\lim _{t \rightarrow \infty}u(1, t).$ Please help.
Hint:
You have formulated the answer to the PDE and it's almost true except that because of constants in it and your equation being linear if there are two answers for this PDE so is any of their linear combination, that's why you need to integrate your answer to attain the most general case: $$u(x,t)=\int_{-\infty}^{\infty}(A(\lambda)\cos(\lambda x)+B(\lambda)\sin(\lambda x))e^{-\lambda^2t}d\lambda$$