Solution:
Given $\lim\limits_{\text{x} \rightarrow0}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{2\text{x}^{2}+\text{x}-3}$
$=\lim\limits_{\text{x} \rightarrow1}\frac{\big(\sqrt{\text{x}}-1\big)\big(2\text{x}-3\big)}{\text{x}\big(2\text{x}+3\big)-1\big(2\text{x}+3\big)}=\lim\limits_{\text{x} \rightarrow 1}\frac{\big(\sqrt{\text{x}}-1\big)(\big(2\text{x}-3\big)}{\big(\text{x}-1\big)\big(2\text{x}+3\big)}$
$=\lim\limits_{\text{x} \rightarrow 1}\frac{2\text{x}-3}{\big(\sqrt{\text{x}}+1\big)\big(2\text{x}+3\big)}$
Taking limit, we get
$=\frac{2(1)-3}{\big(\sqrt{\text{x}}+1\big)\big(2\times1+3\big)}=\frac{-1}{2\times5}=\frac{-1}{10}$
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