Question:
The value of $\displaystyle \lim_{x \to \infty} \frac{x^3 + x^2 - 5}{3x^3 + 7}$ is:
The value of $\displaystyle \lim_{x \to \infty} \frac{x^3 + x^2 - 5}{3x^3 + 7}$ is:
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In $\lim_{x\to\infty}$, keep highest degree terms only.
Updated On: Apr 23, 2026
- 1
- $\frac{1}{3}$
- $-\frac{5}{7}$
- $\frac{5}{7}$
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The Correct Option is B
Solution and Explanation
Concept:
For limits at infinity, divide by highest power.
Step 1: Divide numerator and denominator by $x^3$.
\[ \lim_{x \to \infty} \frac{1 + \frac{1}{x} - \frac{5}{x^3}}{3 + \frac{7}{x^3}} \]
Step 2: Apply limit.
\[ \frac{1 + 0 - 0}{3 + 0} = \frac{1}{3} \]
Hence, the value is $\frac{1{3}$.
Step 1: Divide numerator and denominator by $x^3$.
\[ \lim_{x \to \infty} \frac{1 + \frac{1}{x} - \frac{5}{x^3}}{3 + \frac{7}{x^3}} \]
Step 2: Apply limit.
\[ \frac{1 + 0 - 0}{3 + 0} = \frac{1}{3} \]
Hence, the value is $\frac{1{3}$.
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