g) \( \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x^{2}+x}-\frac{1}{x}\right) \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Mind Expander
To evaluate the limit \( \lim _{x \rightarrow 0^{+}}\left(\frac{1}{x^{2}+x}-\frac{1}{x}\right) \), we start by finding a common denominator for the two fractions: \[ \frac{1}{x^{2}+x}-\frac{1}{x} = \frac{x - (x^{2} + x)}{x(x^{2} + x)} = \frac{x - x^{2} - x}{x(x^{2} + x)} = \frac{-x^{2}}{x(x^{2} + x)} = \frac{-x}{x^{2} + x} \] Next, we simplify this expression: \[ \frac{-x}{x^{2}+x} = \frac{-x}{x(x+1)} = \frac{-1}{x+1} \] Now we take the limit as \(x\) approaches \(0^{+}\): \[ \lim _{x \rightarrow 0^{+}} \frac{-1}{x+1} \] As \(x\) approaches \(0\), \(x+1\) approaches \(1\). Thus, the limit becomes: \[ \lim _{x \rightarrow 0^{+}} \frac{-1}{x+1} = \frac{-1}{0+1} = -1 \] Therefore, the limit is \[ \boxed{-1} \]
