In a C*-algebra $A$, $x$ is self-adjoint iff $\lim_{t\to 0}(1/t)(\Vert 1-itx\Vert-1)=0$. The question I am having trouble with is the following:

Let $A$ be a C$^*$-algebra. Show that an element $x$ of $A$ is self-adjoint iff $\lim_{t\to 0}(1/t)(\Vert 1-itx\Vert-1)=0$. (Hint: If $h\in A$ is self-adjoint, then $\exp(ith)=1+ith+o(t)$ is unitary for every $t\in\mathbb{R}$. If $k$ is another nonzero self-adjoint element of $A$, then $\Vert 1+ith-tk\Vert\geq\Vert 1-tk\Vert\neq 1+o(t)$.)

(In the book it is written $O(t)$ instead of $o(t)$, but this is the correct statement - see comment below)
This is Exercise 4.3 from Takesaki's "Theory of Operator Algebras I". I managed to solve the first implication ($x$ self-adjoint implying the limit being zero) using the hint, but I'm having problems with the converse.
In the hint, I belie he means we should decompose $x$ in the real and imaginary parts, $x=h+ik$, so the real part of $1+itx$ is $1-tk$, and then we work with it to show that $k\neq 0$ gives a contradiction.

So suppose that $\inf \text{Sp}(k) = \alpha$ and $\sup \text{Sp}(k) = \beta$.  If $\beta \ge 0$, then $\|1+tk\| = 1+t\beta$ for sufficiently small $t > 0$.  If $\alpha \le 0$ then $\|1+tk\| = 1+t\alpha$ for sufficiently small $t < 0$.  If $\alpha =\beta = 0$ then $k=0$.  So enumerating all the cases, we see that $\|1+tk\| = 1 + o(t)$ fails as $t \to 0$. 
