Compute the limit: lim (2x2 - 3x + 4)/x + (5x - 4)/x as x approaches 0. Integrate and find an exact simplified value for the integral of (1+cot2 x)sec2 x dx from pi/6 to pi/3.

$$\lim_{x \to 0} \left( \frac{2x^2 - 3x + 5x + 4 - 4}{x} \right)$$ $$\lim_{x \to 0} \left( \frac{2x^2 + 2x}{x} \right)$$

Since $x \to 0$, but $x \neq 0$, we can cancel $x$.

$$\lim_{x \to 0} \left( \frac{x(2x + 2)}{x} \right)$$ $$\lim_{x \to 0} (2x + 2)$$

$$2(0) + 2 = 2$$

The final answer is $\boxed{2}$.

5. Integrate and find an exact simplified value for

$$\int_{\pi/6}^{\pi/3} (1+\cot^2 x)\sec^2 x\,dx$$

Recall the identity $1+\cot^2 x = \csc^2 x$.
So the integrand becomes:

$$\csc^2 x \sec^2 x$$

Now, express $\csc^2 x$ and $\sec^2 x$ in terms of $\sin x$ and $\cos x$:

$$\frac{1}{\sin^2 x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$$

We know that $\sin(2x) = 2\sin x \cos x$. Squaring both sides gives $\sin^2(2x) = 4\sin^2 x \cos^2 x$.
Therefore, $\sin^2 x \cos^2 x = \frac{1}{4}\sin^2(2x)$.
Substitute this back into the integrand:

$$\frac{1}{\frac{1}{4}\sin^2(2x)} = \frac{4}{\sin^2(2x)} = 4\csc^2(2x)$$

We need to find $\int 4\csc^2(2x)\,dx$.
Let $u = 2x$, so $du = 2\,dx$, which means $dx = \frac{1}{2}\,du$.

$$\int 4\csc^2(u) \left(\frac{1}{2}\right)\,du = 2\int \csc^2(u)\,du$$

The integral of $\csc^2 u$ is $-\cot u$.

$$= -2\cot u + C = -2\cot(2x) + C$$

At the upper limit $x = \frac{\pi}{3}$:

$$-2\cot\left(2 \cdot \frac{\pi}{3}\right) = -2\cot\left(\frac{2\pi}{3}\right)$$

We know $\cot\left(\frac{2\pi}{3}\right) = -\frac{1}{\sqrt{3}}$.

$$= -2\left(-\frac{1}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}$$

At the lower limit $x = \frac{\pi}{6}$:

$$-2\cot\left(2 \cdot \frac{\pi}{6}\right) = -2\cot\left(\frac{\pi}{3}\right)$$

We know $\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}$.

$$= -2\left(\frac{1}{\sqrt{3}}\right) = -\frac{2}{\sqrt{3}}$$

$$\frac{2}{\sqrt{3}} - \left(-\frac{2}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}}$$

To rationalize the denominator, multiply by $\frac{\sqrt{3}}{\sqrt{3}}$:

$$\frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$$

The final answer is $\boxed{\frac{4\sqrt{3}}{3}}$.