The infinite series \[{\text{f}}\left( {\text{x}} \right) = {\text{x}} – \frac{{{{\text{x}}^3}}}{{3!}} + \frac{{{{\text{x}}^5}}}{{5!}} – \frac{{{{\text{x}}^7}}}{{7!}}\,…\,\infty \] converges to A. cos(x) B. sin(x) C. sin h(x) D. ex

The correct answer is $\boxed{\text{B}}$. This is the Taylor series for $\sin(x)$ centered at $x=0$.

The Taylor series for a function $f$ centered at $x=0$ is given by

$$f(x) = \sum_{n=0}^{\infty} \frac{{{f}^{(n)}}(0){{x}^{n}}}{{n!}}$$

where $f^{(n)}(x)$ is the $n$th derivative of $f$ evaluated at $x=0$.

In this case, $f(x) = \sin(x)$ and $f^{(n)}(x) = \cos(x)$ for all $n$. Therefore, the Taylor series for $\sin(x)$ centered at $x=0$ is

$$\sin(x) = \sum_{n=0}^{\infty} \frac{{\cos(0){{x}^{n}}}{{n!}} = x – \frac{{{{\text{x}}^3}}}{{3!}} + \frac{{{{\text{x}}^5}}}{{5!}} – \frac{{{{\text{x}}^7}}}{{7!}}\,…\,\infty$$

As you can see, this is the same series as the one given in the question. Therefore, the infinite series converges to $\sin(x)$.