Limit of ((x²+1)/(x²−2)) raised to x²
Evaluate the limit: $$ L = \lim_{x \to \infty} \left(\frac{x^2+1}{x^2-2}\right)^{x^2}. $$
Explanation of Method
As $x\to\infty$, the base $\frac{x^2+1}{x^2-2}\to1$ while the exponent $x^2\to\infty$, giving the indeterminate form $1^\infty$ (unlike the previous problem, where the base tended to $\frac12\neq1$). Taking logarithms converts the power into a product; writing the base as $1+\frac{3}{x^2-2}$ (since the numerator minus the denominator is exactly $3$) makes it clear that $\ln(\text{base})\sim\frac3{x^2-2}$, and multiplying by the exponent $x^2$ gives a ratio that tends to a finite, nonzero constant.
- Rewrite the base as $1$ plus a small correction term.
- Take logarithms and approximate using $\ln(1+u)\approx u$.
- Multiply by the exponent $x^2$ and take the limit.
- Exponentiate to recover $L$.
Step 1 · Rewrite the base as $1+u$
Since $(x^2+1)-(x^2-2)=3$:
$$ \frac{x^2+1}{x^2-2} = \frac{(x^2-2)+3}{x^2-2} = 1+\frac{3}{x^2-2}. $$
As $x\to\infty$, $\dfrac3{x^2-2}\to0$, confirming the base tends to $1$.
Step 2 · Take logarithms and approximate
Let $L=\lim_{x\to\infty}\left(1+\frac3{x^2-2}\right)^{x^2}$. Taking $\ln$:
$$ \ln L = \lim_{x\to\infty} x^2\ln\left(1+\frac3{x^2-2}\right). $$
Since $\frac3{x^2-2}\to0$, use the approximation $\ln(1+u)\approx u$ for small $u$:
$$ \ln\left(1+\frac3{x^2-2}\right) \approx \frac{3}{x^2-2}. $$
Step 3 · Multiply by the exponent and take the limit
Substituting:
$$ \ln L = \lim_{x\to\infty} x^2\cdot\frac{3}{x^2-2} = \lim_{x\to\infty}\frac{3x^2}{x^2-2}. $$
Dividing numerator and denominator by $x^2$:
$$ \frac{3x^2}{x^2-2} = \frac{3}{1-\frac2{x^2}} \longrightarrow \frac{3}{1-0} = 3. $$
So $\ln L = 3$.
Step 4 · Exponentiate
$$ L = e^{\ln L} = e^{3}. $$
Summary Table
| Quantity | Expression | Limit as $x\to\infty$ |
|---|---|---|
| Base | $1+\dfrac3{x^2-2}$ | $1$ |
| $x^2\ln(\text{base})$ | $\dfrac{3x^2}{x^2-2}$ | $3$ |
| $L$ | $e^{\ln L}$ | $e^3$ |
Additional Notes
- Contrast this with the previous problem, $\left(\frac{x-2}{2x+1}\right)^{x^2}$, where the base tended to $\frac12\neq1$: there, no delicate cancellation was needed and the limit was immediately $0$. Here, the base tends to exactly $1$, making this a genuine $1^\infty$ form that requires the log-and-approximate technique to resolve.
- The key algebraic step is rewriting the base as $1+\frac{3}{x^2-2}$: the numerator's excess over the denominator (here, exactly $3$) becomes the coefficient that ultimately appears in the exponent of the final answer.
- This matches the general pattern $\left(1+\frac{c}{g(x)}\right)^{g(x)}\to e^c$ whenever $g(x)\to\infty$, here with $c=3$ and $g(x)=x^2-2\sim x^2$.
Result
The value of the limit is: