Why One-to-Infinity Powers Need Care
A power can have a base approaching \(1\) while its exponent grows without bound. Simply substituting the limiting values suggests the informal form \(1^\infty\), but that form does not determine the limit. The rate at which the base approaches \(1\) matters: different rates can produce different limits.
For instance, powers of the form \(\left(1+\frac{c}{x}\right)^x\), with \(c\) a fixed nonzero real number and \(x\) tending to infinity, approach \(e^c\), not \(1\). The assumption \(c\ne 0\) matters: if \(c=0\), the power is identically \(1\). We will see how logarithms make the role of the rate precise.
The Logarithmic Transformation
Suppose \(B(x)>0\) for all \(x\) sufficiently close to the limit point, \(B(x)\to 1\), and \(G(x)\to+\infty\). The power \(B(x)^{G(x)}\) is positive. Its natural logarithm is \(G(x)\ln B(x)\), so the original limit can be studied through this product.
The finite-limit conclusion follows from continuity of the exponential function: applying \(\exp\) to a quantity tending to \(L\) gives a quantity tending to \(\exp(L)\). The same reasoning with the extended limits uses that \(\exp(t)\to+\infty\) as \(t\to+\infty\), and \(\exp(t)\to0\) as \(t\to-\infty\).
The logarithm changes the question from evaluating an indeterminate power to evaluating a product. That product may be rewritten as a quotient if doing so gives a useful form for L'Hôpital's Rule. The rule applies only when its hypotheses are met: in particular, the quotient must have indeterminate form \(0/0\) or \(\infty/\infty\), the functions must be differentiable near the limit point, and the denominator's derivative must not vanish nearby. It is the logarithmic expression—not the original power—to which the rule is applied.
Worked Example: A Base of the Form One Plus a Constant Over \(x\)
Worked Example: A Base of the Form One Plus a Constant Over \(x\)
Let \(c\ne0\) be fixed. Evaluate \(\displaystyle\lim_{x\to\infty}\left(1+\frac{c}{x}\right)^x\).
For all sufficiently large \(x\), the base \(1+c/x\) is positive. The base tends to \(1\), and the exponent tends to \(+\infty\). Define the logarithmic product \[ x\ln\left(1+\frac{c}{x}\right) = \frac{\ln(1+c/x)}{1/x}. \] As \(x\to\infty\), both numerator and denominator tend to \(0\), so this is a \(0/0\) quotient. Differentiating numerator and denominator gives \[ \frac{\dfrac{-c/x^2}{1+c/x}}{-1/x^2} = \frac{c}{1+c/x} \longrightarrow c. \] Therefore the logarithmic product tends to \(c\), and the original power tends to \(e^c\): \[ \lim_{x\to\infty}\left(1+\frac{c}{x}\right)^x=e^c. \] For example, with \(c=3\), the limit is \(e^3\), which is not \(1\). The condition \(c\ne0\) ensures \(e^c\ne1\).
Check: The logarithm of the power is exactly \(x\ln(1+c/x)\), whose limit was computed as \(c\). Exponentiating that result gives \(e^c\), as required.
Worked Example: A Finite Endpoint
Worked Example: A Finite Endpoint
Evaluate \(\displaystyle\lim_{x\to0^+}(1+x)^{1/x}\).
For \(x>0\) near \(0\), the base \(1+x\) is positive and the exponent \(1/x\) tends to \(+\infty\). Taking the logarithmic product gives \[ \frac{1}{x}\ln(1+x)=\frac{\ln(1+x)}{x}. \] The numerator and denominator both tend to \(0\). L'Hôpital's Rule applies, since the functions are differentiable near \(0\) and the derivative of the denominator is \(1\), which does not vanish. Hence \[ \lim_{x\to0^+}\frac{\ln(1+x)}{x} = \lim_{x\to0^+}\frac{1/(1+x)}{1} =1. \] The logarithmic product tends to \(1\), so the power tends to \[ \lim_{x\to0^+}(1+x)^{1/x}=e^1=e. \]
Check: The logarithm of the proposed limiting value \(e\) is \(1\), matching the limit of \(\ln\bigl((1+x)^{1/x}\bigr)\) just computed.
Why the Logarithm Must Be Evaluated Before Exponentiating
The transformed limit need not be a finite number. A logarithmic product that approaches \(0\) leads to a power approaching \(e^0=1\). A product approaching a negative finite number leads to a limit strictly between \(0\) and \(1\); one approaching a positive finite number leads to a limit greater than \(1\). If the product diverges, its direction determines whether the power tends to \(0\) or grows without bound.
Worked Example: A Power That Tends to One
Evaluate \(\displaystyle\lim_{x\to0^+}(1+x^2)^{1/x}\).
The base is positive near \(0\) and tends to \(1\), while \(1/x\to+\infty\). The logarithmic product is \[ \frac{\ln(1+x^2)}{x}. \] The numerator and denominator both tend to \(0\). Differentiating gives \[ \lim_{x\to0^+}\frac{\ln(1+x^2)}{x} = \lim_{x\to0^+}\frac{2x/(1+x^2)}{1} =0. \] Therefore \[ \lim_{x\to0^+}(1+x^2)^{1/x}=e^0=1. \]
Check: The logarithm of the power tends to \(0\), and exponentiating \(0\) gives \(1\). This does not mean that every power with base tending to \(1\) and exponent tending to infinity tends to \(1\): the preceding example had a logarithmic product tending to \(1\), not \(0\).
Worked Example: A Power That Tends to Zero
Worked Example: A Power That Tends to Zero
Evaluate \(\displaystyle\lim_{x\to\infty}\left(1-\frac{2}{x}\right)^{x^2}\).
The base is positive for \(x>2\), tends to \(1\), and the exponent tends to \(+\infty\). Its logarithmic product is \[ x^2\ln\left(1-\frac{2}{x}\right) = x\left[x\ln\left(1-\frac{2}{x}\right)\right]. \] First evaluate the expression in square brackets by writing it as a quotient: \[ x\ln\left(1-\frac{2}{x}\right) = \frac{\ln(1-2/x)}{1/x}. \] Both numerator and denominator tend to \(0\). Their derivatives give \[ \frac{\dfrac{2/x^2}{1-2/x}}{-1/x^2} = -\frac{2x}{x-2} \longrightarrow -2. \] Thus \(x\ln(1-2/x)\to-2\). Since \(x\to+\infty\), multiplying \(x\) by a quantity tending to \(-2\) shows \[ x^2\ln\left(1-\frac{2}{x}\right)\longrightarrow-\infty. \] The power therefore tends to \(e^{-\infty}=0\): \[ \lim_{x\to\infty}\left(1-\frac{2}{x}\right)^{x^2}=0. \]
Check: The logarithmic product tends to \(-\infty\), so the logarithm of the power becomes arbitrarily negative. Exponentiating such values tends to \(0\), consistent with the result.
A Reliable Procedure
For a power with positive base tending to \(1\) and exponent tending to \(+\infty\), organize the calculation before choosing a differentiation method.
Confirm that the base is positive wherever the power is considered, and that it tends to \(1\).
For base \(B(x)\) and exponent \(G(x)\), calculate \(G(x)\ln B(x)\).
Use algebra first when it simplifies the product. If converting to a quotient, check that it has form \(0/0\) or \(\infty/\infty\) before applying L'Hôpital's Rule.
If it is \(L\), the power tends to \(e^L\). If it is \(+\infty\) or \(-\infty\), the power tends to \(+\infty\) or \(0\), respectively.
Common Mistake: Substituting the Limiting Base
A tempting but incorrect calculation is \[ \left(1+\frac{3}{x}\right)^x \longrightarrow 1^\infty=1. \] This treats the changing base as if it were exactly \(1\) before the exponent is applied. It loses the rate at which the base approaches \(1\). The corrected calculation evaluates \[ x\ln\left(1+\frac{3}{x}\right)\longrightarrow3, \] and therefore gives the limit \(e^3\), not \(1\).
The notation \(1^\infty\) is a warning that the base and exponent limits alone are insufficient; it is not an arithmetic rule for finding the answer. In the examples above, the same general form produced \(e\), \(1\), and \(0\). The logarithmic product distinguished these outcomes by recording how the base's departure from \(1\) combines with the growing exponent.
Why This Method Matters
Variable powers occur whenever both a base and an exponent change, so direct substitution often cannot determine their limits. Logarithms provide an exact transformation, not an approximation: for each positive base, the logarithm of the power equals the exponent times the logarithm of the base. Once that product is evaluated, continuity and the limiting behavior of the exponential function finish the argument.
In particular, L'Hôpital's Rule is not a shortcut to apply to a power merely because its form is indeterminate. It becomes relevant only after the logarithmic transformation and a valid quotient conversion. Keeping these steps separate helps ensure that each rule is used under the conditions that justify it.
Check Your Understanding
For each question, identify the logarithmic product before deciding what the power tends to.
- In the form \(B(x)^{G(x)}\), what conditions allow you to write its logarithm as \(G(x)\ln B(x)\)?
- If \(B(x)\to1\), \(G(x)\to+\infty\), and \(G(x)\ln B(x)\to L\) for a finite \(L\), what is the limit of the power?
- Why does the notation \(1^\infty\) not determine a limit?
- What forms must a quotient have before L'Hôpital's Rule can be applied?
- If the logarithmic product tends to \(-\infty\), what does the original positive power tend to, and why?
