Limit of ((x−2)/(2x+1)) raised to x²
Evaluate the limit: $$ L = \lim_{x \to \infty} \left(\frac{x-2}{2x+1}\right)^{x^2}. $$
Explanation of Method
As $x\to\infty$, the base $\frac{x-2}{2x+1}$ tends to the fixed constant $\frac12$ (found by dividing numerator and denominator by $x$), while the exponent $x^2\to\infty$. Whenever a base converges to a constant strictly between $-1$ and $1$ (here, exactly $\frac12$) and the exponent grows without bound, the whole expression is forced to $0$ — this can be seen quickly by taking logarithms and noting that the leading-order behavior of $\ln(\text{base})$, a nonzero negative constant, completely dominates once multiplied by $x^2\to\infty$.
- Find the limit of the base as $x\to\infty$.
- Take logarithms and identify the leading-order behavior of $\ln L$.
- Conclude that $\ln L\to-\infty$, and hence $L\to0$.
Step 1 · Find the limit of the base
Dividing numerator and denominator by $x$:
$$ \frac{x-2}{2x+1} = \frac{1-\frac2x}{2+\frac1x} \longrightarrow \frac{1-0}{2+0} = \frac12 \qquad \text{as } x\to\infty. $$
Step 2 · Take logarithms
Let $L=\lim_{x\to\infty}\left(\frac{x-2}{2x+1}\right)^{x^2}$. Taking $\ln$ (formally, before justifying the limit exists):
$$ \ln L = \lim_{x\to\infty} x^2\ln\left(\frac{x-2}{2x+1}\right). $$
Since the base tends to the fixed constant $\frac12$ (Step 1), and $\ln$ is continuous at $\frac12$:
$$ \ln\left(\frac{x-2}{2x+1}\right) \longrightarrow \ln\frac12 = -\ln2, \qquad \text{a fixed negative constant.} $$
Step 3 · Conclude $\ln L\to-\infty$
Since $\ln\left(\frac{x-2}{2x+1}\right)$ tends to the nonzero negative constant $-\ln2$ while $x^2\to+\infty$, the product tends to $-\infty$:
$$ \ln L = \lim_{x\to\infty} x^2\cdot\underbrace{\ln\left(\frac{x-2}{2x+1}\right)}_{\to\,-\ln2\,<\,0} = -\infty. $$
Exponentiating:
$$ L = e^{\ln L} = e^{-\infty} = 0. $$
Summary Table
| Quantity | Limit as $x\to\infty$ |
|---|---|
| Base $\dfrac{x-2}{2x+1}$ | $\dfrac12$ |
| $\ln\left(\dfrac{x-2}{2x+1}\right)$ | $-\ln2$ (nonzero, negative) |
| $\ln L = x^2\cdot(\to-\ln2)$ | $-\infty$ |
| $L$ | $0$ |
Additional Notes
- Because the base tends to a fixed constant with absolute value strictly less than $1$, this is not a $1^\infty$ indeterminate form at all — no delicate cancellation is needed, since a fixed base less than $1$ in magnitude raised to an unboundedly growing power always tends to $0$, regardless of the precise rate at which the base approaches its limiting value.
- It is worth checking the type of indeterminate form before reaching for a detailed expansion: bases tending to $1$ require the careful $(1+u)^{1/u}\to e$ machinery (as in several other limits on this site), but bases tending to any other constant inside $(-1,1)$ settle the question immediately.
- Had the base instead tended to a value greater than $1$ in absolute value (for example, if the coefficients were swapped to give a base tending to $2$), the same reasoning would show $L\to+\infty$ instead, since $\ln(\text{base})$ would then be a positive constant.
Result
The value of the limit is: