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**VitaX** Test for convergence:

$\displaystyle \int_4^{\infty} \frac{1}{\sqrt{x} - 1} dx$

$\displaystyle 0\leq \frac{1}{\sqrt{x}} \leq \frac{1}{\sqrt{x} - 1}$

$\displaystyle \lim_{b \rightarrow \infty} \int_4^b \frac{1}{\sqrt{x}} dx$

$\displaystyle \lim_{b \rightarrow \infty} \int_4^b x^{-\frac{1}{2}}$

$\displaystyle \lim_{b \rightarrow \infty} 2x^{\frac{1}{2}}|_4^b$

$\displaystyle \lim_{b \rightarrow \infty} \left[2(b)^{\frac{1}{2}} - 2(4)^{\frac{1}{2}}\right] = -4$

**The final answer for the limit is wrong.**

Since the limit is finite, the function converges. Did I do this one right?

**No**

$\displaystyle \int_{\pi}^{\infty} \frac{1 + sinx}{x^2} dx$

How would I go about solving this one? Which function would I choose to integrate to see if a finite value is found?

**Since** $\displaystyle \color{red} sin(x) \leq 1 \, \forall \, x \in R \,\$ **then** $\displaystyle \,\ \color{red} \frac{1+sin(x)}{x^2} \leq \frac{2}{x^2}$

**So .. ?**