Since the denominator will stay positive but go to $0$, and the numerator will be a positive number different from $0$, the quotient will always be positive.
As $x$ gets closer and closer to $\frac{1}{2}$, the denominator will get smaller and smaller and smaller. Dividing a number which will be very close to $10.5$ by a number that is getting smaller, and smaller, and smaller, and smaller (but always stays positive) will result in a number that is getting larger, and larger, and larger, and larger. The quotient will grow without bound and stay positive. So $\lim_{x\to\frac{1}{2}^+}\frac{10+x}{(2x-1)^3} = \infty.$
For a basic calculus course, I would accept that as a valid answer; your instructor, though, may have different standards. Exactly how to write it so as to satisfy her will depend on what those standards are, and only she can tell you that.
If you need this to be done using $\epsilon$-$\delta$ type arguments, you need to show that for any $N\gt 0$ you can find $\delta\gt 0$ (which may depend on $N$) with the property that any $x$ such that $0\lt x-\frac{1}{2}\lt \delta$ will satisfy $\frac{10+x}{(2x-1)^3}\gt N.$ How to figure it out? Note that for any such $x$, we will have $10+x\gt 10$. So it is enough to make sure that $\frac{10.5}{(2x-1)^3}$ is larger than $N$. To make this larger than $N$, we need to make sure that $\frac{1}{(2x-1)^3} \gt \frac{N}{10}.$ To make sure this is true, it is enough to make sure that $(2x-1)^3 \lt \frac{10}{N}$ which requires $2x-1 \lt \sqrt[3]{\frac{10}{N}}$ which requires $2x \lt 1 + \sqrt[3]{\frac{10}{N}}$ which requires $x\lt \frac{1}{2}+\frac{1}{2}\sqrt[3]{\frac{10}{N}}.$ So, let us set $\delta = \frac{1}{2}\sqrt[3]{\frac{10}{N}}$.
If $0\lt x-\frac{1}{2}\lt \delta = \frac{1}{2}\sqrt[3]{\frac{10}{N}},$ then $0 \lt 2x - 1 \lt \sqrt[3]{\frac{10}{N}}$ so $0\lt \frac{1}{\sqrt[3]{\frac{10}{N}}} \lt \frac{1}{2x-1},$ which is the same as $0 \lt \frac{\sqrt[3]{N}}{\sqrt[3]{10}}\lt \frac{1}{2x-1};$ hence $0\lt \frac{N}{10} \lt \frac{1}{(2x-1)^3}$ which means that we will have $0\lt N \lt \frac{10}{(2x-1)^3} \lt \frac{10+x}{(2x-1)^3}.$ Thus, for any $N\gt 0$ there exists $\delta\gt0$ (for example, $\delta = \frac{1}{2}\sqrt[3]{\frac{10}{N}}$, though any smaller $\delta$ will also work), such that if $0\lt x-\frac{1}{2}\lt \delta$, then $\frac{10+x}{(2x-1)^3} \gt N$. This proves that $\lim_{x\to\frac{1}{2}^+}\frac{10+x}{(2x-1)^3} = \infty.$