Is it a new type of induction? (Infinitesimal induction) Is this even true? Suppose we want to prove Euler's Formula with induction for all positive real numbers. 
At first this seems baffling, but an idea struck my mind today. 


Prove: $$e^{ix}=\cos x+i\sin x \ \ \ \forall \ \ x\geq0$$

For $x=0$, we have $$1=1$$
So the equality holds. 
Now let us assume that the given equality holds for some $x=k$. 
$$e^{ik}=\cos k+i\sin k$$
Now, this is where I added my "own" axiom. Please answer whether this "axiom" is true or not. Now this equality must hold for $x=k+\Delta k$ also, for some infinitely small positive change $\Delta k$ (infinitesimal).

So $e^{i(k+\Delta k)}=e^{ik}.e^{i\Delta k}=(\cos k+i\sin k)(\cos\Delta k+i\sin\Delta k)$
 $$=\cos k\cos\Delta k-\sin k\sin\Delta k+i\sin k\cos\Delta k+i\cos k\sin \Delta k$$
$$=\cos(k+\Delta k)+i\sin(k+\Delta k)$$
So we proved it for $x=k+\Delta k$, and hence it must hold for all $x\geq0$. 
So is this approach correct, and is this a new type of induction?

Most of the comments below are indicating that the proof above is wrong. But we see the Euler's Formula does hold for all $x\geq0$, so can someone give a good counter example where this proof doesn't work. Or in other words, these statements are used to prove a wrong equality.?

Update: Okay, some comments below are suggesting that Euler's Formula is definitely true. So if we prove somehow that $$e^{i\Delta k}=\cos\Delta k+i\sin\Delta k$$ then? But how can we prove it for infinitesimal? Will the concept of limits be used? Can someone solve this mystery?

While the argument in general isn't true, as described in other answers, there is another way how to prove euality of functions. The important condition is that both $e^{ix}$ and $\cos x + i sin x$ are holomorphic functions. The Identity theorem states that

If two holomorphic functions $f$ and $g$ on a domain $D$ agree on a set $S$ which has an accumulation point in $D$ then $f = g$ on all of $D$.

So if you prove that there is a sequence of points $x_i$ such that $lim_{i\to\infty} x_i\in D$, you can use the theorem to prove the equality on the whole domain. And for this you could use the induction principle on $i$.
For example you could set $x_0=0$, $x_1=1$ and $x_n=(x_{n-1}+x_{0})/2$. Then if you prove that the equality holds for $0$, $1$ and that if it holds for $x$ and $y$ it also holds for $(x+y)/2$, the theorem gives you equality everywhere.
