$lim_{x\rightarrow0}\frac{x^{2}-3x}{xsqrt{x}}$
$lim_{x\rightarrow0}\frac{x^{2}-3x}{xsqrt{x}}$
Answer
Explanation:
Step1: Simplify the expression
First, factor out (x) from the numerator: (\frac{x^{2}-3x}{x\sqrt{x}}=\frac{x(x - 3)}{x\sqrt{x}}). Then, cancel out the common - factor (x) (since (x\to0) but (x\neq0) during the limit - taking process), getting (\frac{x - 3}{\sqrt{x}}).
Step2: Analyze the limit
As (x\to0^{+}), (\lim_{x\to0^{+}}\frac{x - 3}{\sqrt{x}}=\lim_{x\to0^{+}}\left(\frac{x}{\sqrt{x}}-\frac{3}{\sqrt{x}}\right)=\lim_{x\to0^{+}}\left(\sqrt{x}-\frac{3}{\sqrt{x}}\right)). (\lim_{x\to0^{+}}\sqrt{x}=0) and (\lim_{x\to0^{+}}\frac{3}{\sqrt{x}}=+\infty). So, (\lim_{x\to0^{+}}\left(\sqrt{x}-\frac{3}{\sqrt{x}}\right)=-\infty). As (x\to0^{-}), (\sqrt{x}) is not a real - number in the set of real numbers. So, we consider the right - hand limit.
Answer:
(-\infty)