If $|y_m| \to 0$ then there exists $k_m$ such that $k_m |y_m| \to t$

Lemma 3.43 Suppose $B_m$ are elements of $G$ and that $B_m \to I$. Let $Y_m = \log B_m$, which is defined for all sufficiently large $m$. Suppose that $Y_m$ is nonzero for all $m$ and that $Y_m/\|Y_m\| \to Y \in \operatorname{M}_n(\mathbb{C})$. Then $Y$ is in $\mathfrak{g}$.

Proof. For any $t \in \mathbb{R}$, we have $(t/\|Y_m\|)Y_m \to t Y$. Note that since $B_m \to I$, we have $\|Y_m\| \to 0$. Thus, we can find integers $k_m$ such that $k_m \|Y_m\| \to t$. We have, then, $$ e^{k_m Y_m} = \exp \left[ (k_m \|Y_m\|) \frac{Y_m}{\|Y_m\|} \right] \to e^{tY}.$$

(Original image here.)

The group $G$ is a Lie group of matrices. This was taken from Brian Hall. It’s reallly just analysis question. But I’m not entirely sure why the $k_m$’s exists