Derivative of distance along a smooth curve I am struggling to solve the following problem from 'Introduction to Riemannian Manifolds' by John M. Lee

$(M,g)$ be a connected Riemannian manifold. $\gamma:(-\epsilon,\epsilon)\to M$ be a smooth curve then $$\lim_{t\to0}\frac{d_g(\gamma(0),\gamma(t))}{t}=|\dot\gamma(0)|_g$$

I was able to show that $$\lim_{t\to0}\frac{d_g(\gamma(0),\gamma(t))}{t}\le \lim_{t\to0+}\frac{l(\gamma|_{[0,t]})}{t}=\lim_{t\to0-}\frac{l(\gamma|_{[t,0]})}{t}=|\dot\gamma(0)|_g$$
Here $l(\cdot)$ denotes length of the curve.
Then I tried to contradict assuming  $$\lim_{t\to0}\frac{d_g(\gamma(0),\gamma(t))}{t}>|\dot\gamma(0)|_g$$
but failed to complete the argument.
Am I doing anything wrong? Please help. Some hints would be highly appreciated.
 A: There is a severe mistake in this question, I misinterpreted the notation in the book due to my ignorance. It was after sir Ted Shifrin's comment that i noticed it (thanks).
The actual question is the following:

$(M,g)$ be a connected Riemannian manifold. $\gamma:(-\epsilon,\epsilon)\to M$ be a smooth curve then $$\lim_{t\to0+}\frac{d_g(\gamma(0),\gamma(t))}{t}=|\dot\gamma(0)|_g$$

The limit doesn't necessarily exist for $t\to0$ please follow the above discussion with sir Ted Shifrin for justification. Also one should notice
$$|\dot\gamma(0)|_g=\lim_{t\to0+}\frac{l(\gamma|_{[0,t]})}{t}\not=\lim_{t\to0-}\frac{l(\gamma|_{[t,0]})}{t}=-|\dot\gamma(0)|_g$$

Now here is the proof that works, which I was able to figure out only after Ted mentioned normal coordinates in the comment--
For $t$ small enough in a normal neighbourhood around $\gamma(0)$ we have
$$\gamma(t)=(x^1(t),\dots,x^n(t))$$
Now $$d(\gamma(0),\gamma(t))=\sqrt{\sum (x^i(t))^2}$$
Therefore
$$\lim_{t\to0+}\frac{d_g(\gamma(0),\gamma(t))}{t}=\lim_{t\to0+}\frac{\sqrt{\sum (x^i(t))^2}}{t}=\lim_{t\to0+}\sqrt{\sum \left(\frac{x^i(t)}{t}\right)^2}=\sqrt{\sum (\dot x^i(0))^2}=|\dot\gamma(0)|_g$$
