Three sharp Depth Lemma inequalities

The maximal ideal of $R$ is $(t)$. Multiplication by $t$ is injective on $R$, so $(t)$ is an $R$-regular sequence of length one. For any nonzero $r\in (t)$, write $r=t^{m}u$ with $m\ge 1$ and $u$ a unit, using the first nonzero coefficient of the power series. In $R\mathrm{/}(r)$, the class of $t^{m-1}$ is nonzero and is annihilated by every element of $(t)$. Thus no regular sequence in $(t)$ has length two, and $\mathrm{depth}R=1$. The same annihilation gives ${\mathrm{depth}}_{R}k=0$.

The first sequence is split exact. The second and third sequences are exact because $t$ is a nonzerodivisor and its cokernel is $k$. Every element of $(t)$ annihilates the nonzero element $(1,0)$ of $k\oplus R$, so this middle module has depth zero.

Substitution of the depth triples $(0,0,1)$ and $(1,1,0)$ into the three bounds of The three Depth Lemma inequalities gives the displayed equalities, so no lower bound can be increased in general.