I posted this question on Math Stack Exchange, but nobody answered so I decided to ask this question here.  

Suppose that $M$ is smooth compact manifold and let $y \to x$. Let also $f \in C^{\infty}(M)$ be a smooth function. I consider the expression $\exp\_y^{-1}(x)(f)$: then it follows that it converges to $\exp\_x^{-1}(x)(f)$. Since $\exp\_x(0)=x$ then $\exp\_x^{-1}(x)=0\_{T\_xM}$ (we are in the tangent space to $M$ at $x$, namely $T\_xM$). Each tangent vector $v$ acts on $f$ via $v \cdot f=\frac{d}{dt}(f \circ \gamma)(0)$ where $\gamma$ is such that $[\gamma]=v$ and $\gamma(0)=x$. In particular to zero vector in tangent space corresponds a constant curve $\gamma(t) \equiv x$ so $f \circ \gamma$ is constant so $\exp\_y^{-1}(x)(f) \to 0$. Is it correct? I found in some book that author claims that it will converge to $f(x)$



Exponential map and convergence
You are right: $\exp:TM\supset U \to M\times M$ is a diffeomorphism onto a neighborhood of the diagonal, thus $\lim\_{y\to x}\exp\_y^{-1}(x) = \exp\_x^{-1}(x) = 0\_x$ in $TM$.


