$\lim _{x \rightarrow-\infty}\left(7^x\right)\left(x^3+5 x^2-8\right)=?$

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02-11-2026
Mathematics

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$\lim _{x \rightarrow-\infty}\left(7^x\right)\left(x^3+5 x^2-8\right)=?$

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-02T16:26:50-05:00

Recognize the indeterminate form $0 \cdot (-\infty)$ as $x \to -\infty$.
Substitute $u = -x$ to rewrite the expression as $\lim_{u \to \infty} \frac{-u^3+5u^2-8}{7^u}$.
Use L'Hôpital's Rule three times to reduce the numerator to a constant.
Conclude that the limit approaches $\boxed{0}$ because exponential growth dominates polynomial growth.

Explanation

Analyzing the Indeterminate Form

To evaluate the limit $\lim _{x \rightarrow-\infty}\left(7^x\right)\left(x^3+5 x^2-8\right)$, we first observe the behavior of the individual factors as $x \rightarrow -\infty$. The exponential term $7^x$ approaches $0$ since the base $7 > 1$. The polynomial term $x^3+5 x^2-8$ behaves like its leading term $x^3$, which approaches $-\infty$. This results in an indeterminate form of type $0 \cdot (-\infty)$.

Transformation and Applying L'Hôpital's Rule

To handle this form, we perform a substitution $u = -x$. As $x \rightarrow -\infty$, $u \rightarrow \infty$. Substituting $x = -u$ into the original limit, we obtain:
\[ \lim _{u \rightarrow \infty} (7^{-u})((-u)^3 + 5(-u)^2 - 8) = \lim _{u \rightarrow \infty} \frac{-u^3 + 5u^2 - 8}{7^u} \]
This transformation converts the expression into a ratio of a polynomial to an exponential function, resulting in an indeterminate form of $\frac{-\infty}{\infty}$, which allows us to apply L'Hôpital's Rule.

Evaluating the Limit using Repeated Derivatives

Applying L'Hôpital's Rule involves differentiating the numerator and the denominator with respect to $u$. We repeat this process until the polynomial in the numerator becomes a constant:

First application: $\lim _{u \rightarrow \infty} \frac{\frac{d}{du}(-u^3 + 5u^2 - 8)}{\frac{d}{du}(7^u)} = \lim _{u \rightarrow \infty} \frac{-3u^2 + 10u}{7^u \ln(7)}$
Second application: $\lim _{u \rightarrow \infty} \frac{\frac{d}{du}(-3u^2 + 10u)}{\frac{d}{du}(7^u \ln(7))} = \lim _{u \rightarrow \infty} \frac{-6u + 10}{7^u (\ln 7)^2}$
Third application: $\lim _{u \rightarrow \infty} \frac{\frac{d}{du}(-6u + 10)}{\frac{d}{du}(7^u (\ln 7)^2)} = \lim _{u \rightarrow \infty} \frac{-6}{7^u (\ln 7)^3}$

Final Conclusion

As $u \rightarrow \infty$, the constant $-6$ remains fixed, while the denominator $7^u (\ln 7)^3$ grows without bound (approaches $\infty$). Therefore, the fraction approaches $0$. Thus, the original limit is $\boxed{0}$.

Examples

In physics and engineering, the decay of certain processes, such as the cooling of an object or the discharge of a capacitor, often involves exponential decay functions multiplied by polynomial growth terms. Understanding how these functions behave as time approaches infinity allows engineers to predict the long-term state of a system, ensuring that the system eventually stabilizes toward a steady-state value, such as absolute zero or zero charge. In this problem, we analyzed how an exponential term dominates a polynomial one, which is fundamental in stability analysis.