Infinite Horizon Transversality Condition
Your confusion comes from the fact that you treat the Transversality condition as a constraint, while it is a condition for optimality. So the formulation at the end of your question is wrong. What you do is you solve your model as usual, and then check if the solution (here the balanced growth path) satisfies the Transversality condition.

The Transversality condition (TVC) is

$$\lim_{t \rightarrow \infty} \beta^t u'(c_t)k_{t+1} =0$$

In your model, $u'(c_t) = c_t^{-\gamma}$ and $c_t = (2-\delta)k_t - k_{t+1}$

So the TVC becomes indeed

$$\lim_{t \rightarrow \infty} \frac {\beta^tk_{t+1}}{[(2-\delta)k_t - k_{t+1}]^{\gamma}} =0$$

What we want is to check whether the (unique) balanced growth path satisfies the Transversality condition. On the balanced growth path we have $k_{t+1} = (1+g)k_t$ where $g$ is a constant determined by the exogenous parameters of the model and the optimizing conditions. So when on the balanced growth path, the TVC becomes

$$\lim_{t \rightarrow \infty} \frac {\beta^t(1+g)k_t}{[(2-\delta)k_t - (1+g)k_t]^{\gamma}} = \lim_{t \rightarrow \infty}\frac {(1+g)}{(1-\delta-g)^{\gamma}}\beta^t k_t^{1-\gamma}=0$$

The constants are unimportant, so we want that

$$\lim_{t \rightarrow \infty}\beta^t k_t^{1-\gamma}=0$$

holds. Note that if $\gamma \geq 1$ (which is the consensus in the literature) the condition does hold, since capital goes in the denominator, and we're done. 

For the case $\gamma &lt;1$, we want the expression to hold at the limit only, i.e. eventually . "Eventually" implies that, if $k^*_s$ is the level of capital at time $s&lt;t$ when the economy arrives at the balanced growth path, we will have $k_t = k_s^*\cdot (1+g)^{t-s}$. Then the TVC becomes

$$\lim_{t \rightarrow \infty}\beta^t \big[k_s^*\cdot (1+g)^{t-s}\big]^{1-\gamma}=0$$

Again, shed things that do not depend on $t$, to get

$$\lim_{t \rightarrow \infty}\beta^t (1+g)^{(1-\gamma)t}=0 \implies \lim_{t \rightarrow \infty}\big[\beta (1+g)^{(1-\gamma)}\big]^t=0$$

So the TVC will hold if and only if

$$\beta (1+g)^{(1-\gamma)} &lt;1$$

In the other question the OP has obtained

$$1+g = [\beta(\alpha+1-\delta)]^{\frac{1}{\gamma}}$$

Substituting, we want

$$\beta \Big([\beta(\alpha+1-\delta)]^{1/\gamma}\Big)^{(1-\gamma)} &lt;1$$

This can be simplified a bit, and it is a restriction on the parameters that must hold in order for the balanced growth path to satisfy the Transversality condition (when $\gamma &lt;1$). For baseline values of the other parameters one can see that the condition is not likely to be satisfied.