Question
\( \lim _{x \rightarrow-=} \frac{x^{2}-x^{2}+x}{x^{3}-x+5} \)
Ask by Bond Moreno. in the United States
Feb 26,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The limit is 0.
Solution
Calculate the limit \( \lim_{x\rightarrow -\infty} \frac{x^{2}-x^{2}+x}{x^{3}-x+5} \).
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{x\rightarrow -\infty}\left(\frac{x^{2}-x^{2}+x}{x^{3}-x+5}\right)\)
- step1: Calculate:
\(\lim _{x\rightarrow -\infty}\left(\frac{x}{x^{3}-x+5}\right)\)
- step2: Rewrite the expression:
\(\lim _{x\rightarrow -\infty}\left(\frac{x}{x^{3}}\times \frac{1}{1-\frac{1}{x^{2}}+\frac{5}{x^{3}}}\right)\)
- step3: Calculate:
\(\lim _{x\rightarrow -\infty}\left(\frac{1}{x^{2}}\times \frac{1}{1-\frac{1}{x^{2}}+\frac{5}{x^{3}}}\right)\)
- step4: Rewrite the expression:
\(\lim _{x\rightarrow -\infty}\left(\frac{1}{x^{2}}\right)\times \lim _{x\rightarrow -\infty}\left(\frac{1}{1-\frac{1}{x^{2}}+\frac{5}{x^{3}}}\right)\)
- step5: Calculate:
\(\lim _{x\rightarrow -\infty}\left(\frac{1}{x^{2}}\right)\times \frac{\lim _{x\rightarrow -\infty}\left(1\right)}{\lim _{x\rightarrow -\infty}\left(1-\frac{1}{x^{2}}+\frac{5}{x^{3}}\right)}\)
- step6: Rewrite the expression:
\(\lim _{x\rightarrow -\infty}\left(\frac{1}{x^{2}}\right)\times \frac{1}{\lim _{x\rightarrow -\infty}\left(1-\frac{1}{x^{2}}+\frac{5}{x^{3}}\right)}\)
- step7: Rewrite the expression:
\(\lim _{x\rightarrow -\infty}\left(\frac{1}{x^{2}}\right)\times \frac{1}{1}\)
- step8: Rewrite the expression:
\(0\times \frac{1}{1}\)
- step9: Calculate:
\(0\times 1\)
- step10: Calculate:
\(0\)
The limit of the expression \( \frac{x^{2}-x^{2}+x}{x^{3}-x+5} \) as \( x \) approaches negative infinity is 0.
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Extra Insights
To evaluate the limit \( \lim _{x \rightarrow 0^{-}} \frac{x^{2}-x^{2}+x}{x^{3}-x+5} \), we first simplify the expression. The numerator simplifies to \( x \) because \( x^{2}-x^{2} = 0 \). Therefore, the limit becomes: \[ \lim _{x \rightarrow 0^{-}} \frac{x}{x^{3}-x+5} \] Now, we can substitute \( x = 0 \) in the denominator: \[ x^{3} - x + 5 = 0^{3} - 0 + 5 = 5 \] Thus, the limit now looks like: \[ \lim _{x \rightarrow 0^{-}} \frac{x}{5} = \frac{0}{5} = 0 \] Therefore, the final result is: \[ \lim _{x \rightarrow 0^{-}} \frac{x^{2}-x^{2}+x}{x^{3}-x+5} = 0 \]
