No I was asking to take it as an example at first we had 2cos and the 2 had gone to the denominator of h now what if we only have cos

To use the standard limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$, we need the denominator to be $\frac{h}{2}$. We currently have $h$.
We can rewrite $h$ as $2 \times \frac{h}{2}$.
So, $\frac{\sin\left(\frac{h}{2}\right)}{h} = \frac{\sin\left(\frac{h}{2}\right)}{2 \times \frac{h}{2}} = \frac{1}{2} \times \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$.

As $h \to 0$:

• $\cos\left(x + \frac{h}{2}\right) \to \cos(x + 0) = \cos x$
• $\frac{1}{2}$ remains $\frac{1}{2}$
• $\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \to 1$ (using the standard limit where $\theta = \frac{h}{2}$)

So, if there were no '2' in the numerator initially, the derivative would be $\boxed{\frac{1}{2} \cos x}$. The '2' in the original problem was essential because it cancelled out the '2' that we introduced in the denominator to match the $\sin\theta/\theta$ form. The cosine term itself doesn't need a special denominator because its limit can be found by direct substitution.