Is it true that:


Let $R$ be a local ring and $\dim R= d$. If $b\subset a$ be two proper ideals of $R$ then for $ n\in {\Bbb N}$, $\varinjlim Ext^d\_R(a^n/b^n,R)=0$



lim Ext(a^n/b^n,R)=0
I prove your question.
Since the short exact sequence
$$0 \to \mathfrak{a}^n/\mathfrak{b}^n \to R/\mathfrak{b}^n \to R/\mathfrak{a}^n \to 0$$
we have the following exact sequence
$$\cdots \to \mathrm{Ext}^d\_R(R/\mathfrak{a}^n,R) \to \mathrm{Ext}^d\_R(R/\mathfrak{b}^n,R) \to \mathrm{Ext}^d\_R(\mathfrak{a}^n/\mathfrak{b}^n,R) \to \cdots $$
Passing to the limit we have the exact sequence


$$\cdots \to H^d\_{\mathfrak{a}}(R) \to H^d\_{\mathfrak{b}}(R) \to \lim \mathrm{Ext}^d\_R(\mathfrak{a}^n/\mathfrak{b}^n,R) \to H^{d+1}\_{\mathfrak{a}}(R) \cong 0$$


by the Grothendieck vanishing theorem. So it is sufficient to show that if $\mathfrak{b} \subseteq \mathfrak{a}$ then  

$$H^d\_{\mathfrak{a}}(R) \to H^d\_{\mathfrak{b}}(R) $$
is surjecitve. Of course, we may assume that $\mathfrak{a} = \mathfrak{b} + (x)$ for some element $x \in \mathfrak{m}$. By [Brodmann-Sharp: local cohomology, Proposition 8.1.2] we have that


$$\cdots \to H^d\_{\mathfrak{a}}(R) \to H^d\_{\mathfrak{b}}(R) \to H^d\_{\mathfrak{b}}(R\_x)$$


Now the claim follows from the fact $\dim R\_x < d$ and the Grothendieck vanishing theorem.


