MCQ
If x is very large, then $\frac{2\text{x}}{1+\text{x}}\text{is:}$
- AClose to 0
- BArbitrarily large
- CLie between 2 and 3
- DClose to 2
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$\lim\limits_{\text{x} \rightarrow0}\frac{\text{x}^{\text{m}}-1}{\text{x}^{\text{n}}-1}$ is equal to:
$1$
$\frac{\text{m}}{\text{n}}$
$\frac{-\text{m}}{\text{n}}$
$\text{m}^{2}\text{n}^{2}$