Why Indeterminate Powers Need a New Approach
For a quotient that has an indeterminate form, L'Hôpital's Rule may provide a way to evaluate the limit. Powers such as \(f(x)^{g(x)}\) can also have indeterminate forms, but they are not quotients. The useful first step is to take a logarithm: this changes the power into a product, which can sometimes be rewritten as a quotient.
The forms considered here are \(0^0\), \(1^\infty\), and \(\infty^0\). These labels describe the separate limits of the base and exponent; they do not determine the limit of the power. The same form can lead to different answers depending on how quickly the base and exponent change.
The Logarithmic Transformation
Suppose \(f(x)>0\) near the point or in the limiting region being considered. This positivity condition allows us to take the natural logarithm, even when \(f(x)\) approaches zero. Let \(y(x)=f(x)^{g(x)}\). The logarithm and exponential are inverse functions, so
Thus, if the transformed expression \(g(x)\ln f(x)\) approaches a limit \(L\), continuity of the exponential function gives \(y(x)\to e^L\). This includes infinite transformed limits: if the product approaches \(-\infty\), then \(y(x)\to0\); if it approaches \(+\infty\), then \(y(x)\to+\infty\).
When the transformed expression is a product, it may be possible to rewrite it as a quotient in the \(0/0\) or \(\infty/\infty\) form before using L'Hôpital's Rule. The hypotheses of that rule still matter: the quotient must have the required indeterminate form, the numerator and denominator must be differentiable in the relevant interval, and the derivative of the denominator must be nonzero there. Taking logarithms does not make L'Hôpital's Rule automatically applicable.
Why the Method Works
The method follows from exact identities, not an approximation. For every positive number \(f(x)\), the logarithm of the power is exactly \(g(x)\ln f(x)\). Therefore the original power is exactly the exponential of that product. If the product approaches \(L\), continuity of the exponential function transfers the limit to the power.
The identity \(\ln(f^g)=g\ln f\) applies here because the base is positive. Once the limit of the logarithm is known, exponentiating gives the original limit. This explains both the procedure and its essential domain condition.
Worked Examples
Worked Example: A Zero-to-Zero Power
Evaluate \(\displaystyle\lim_{x\to0^+}x^x\). As \(x\to0^+\), the base approaches \(0\) and the exponent approaches \(0\), so the form is \(0^0\). For every \(x>0\), the base is positive. Set \(y=x^x\) and take logarithms:
As \(x\to0^+\), this quotient has the form \((-\infty)/(\infty)\). The functions are differentiable for \(x>0\), and the derivative of the denominator is nonzero there, so apply L'Hôpital's Rule:
Therefore \(\ln y\to0\), and exponentiating gives
Check: the transformed logarithm tends to \(0\), and the exponential of that limit is \(1\). Thus the power limit agrees with the exact relation \(y=\exp(\ln y)\).
Worked Example: A One-to-Infinity Power
Evaluate \(\displaystyle\lim_{x\to\infty}\left(1+\frac1x\right)^x\). The base approaches \(1\) and the exponent approaches \(+\infty\), giving the form \(1^\infty\). The base is positive for \(x>0\). Set \(y=(1+1/x)^x\). Then
Put \(t=1/x\). As \(x\to\infty\), \(t\to0^+\), and the quotient becomes \(\ln(1+t)/t\), a \(0/0\) form. The numerator and denominator are differentiable near \(t=0\), with denominator derivative \(1\), so L'Hôpital's Rule gives
Hence \(\ln y\to1\), and the original limit is
Check: the logarithm of the power tends to \(1\), so exponentiating gives \(e\), as required by \(y=\exp(\ln y)\). In particular, a base tending to \(1\) does not force the power to tend to \(1\).
Worked Example: An Infinity-to-Zero Power
Evaluate \(\displaystyle\lim_{x\to\infty}(x+1)^{1/\ln x}\). For \(x>1\), the base is positive and \(\ln x>0\). The base approaches \(+\infty\), while the exponent approaches \(0\), so this is an \(\infty^0\) form. Let \(y=(x+1)^{1/\ln x}\). Its logarithm is
As \(x\to\infty\), both numerator and denominator approach \(+\infty\). Their derivatives are \(1/(x+1)\) and \(1/x\), and the denominator derivative is nonzero for \(x>1\). L'Hôpital's Rule gives
Therefore \(\ln y\to1\), and
Check: substituting the transformed limit into \(y=\exp(\ln y)\) gives \(e^1=e\). The result also shows why the exponent approaching zero is not enough to conclude that the power approaches \(1\).
Different Outcomes from the Same Form
Calling a power indeterminate means that the separate limits of its base and exponent do not settle the limit of the power. It does not mean that the limit fails to exist. For example, consider \(x^{-1/\sqrt{\ln x}}\) as \(x\to\infty\), where \(x>1\). Its base tends to infinity and its exponent tends to zero. Its logarithm is
Exponentiating shows that this power tends to \(0\). By contrast, the previous example, also of the form \(\infty^0\), tended to \(e\). The transformed logarithm determines which outcome occurs.
Common Mistake: Judging the Power by Its Base
A tempting but incorrect argument for \(\left(1+\frac1x\right)^x\) is: “The base tends to \(1\), so the power tends to \(1\).” That treats a changing base as though it were exactly \(1\) before raising it to a changing exponent. The calculation from the worked example shows otherwise:
The correction is to transform the whole power, not just take the limit of its base. The exponent can magnify the effect of a small difference between the base and \(1\). The same caution applies to the other indeterminate forms: neither the base nor the exponent alone determines the answer.
A Reliable Procedure
Confirm that the base is positive in the region where the limit is taken, so its real logarithm is defined.
Write \(\ln y=g(x)\ln f(x)\) for \(y=f(x)^{g(x)}\).
Use algebra first; if rewriting as a quotient produces \(0/0\) or \(\infty/\infty\), apply L'Hôpital's Rule only when its conditions hold.
If the transformed limit is \(L\), the original limit is \(e^L\). A transformed limit of \(-\infty\) gives \(0\), and \(+\infty\) gives \(+\infty\).
Confirm that the result follows from the limit of the logarithm and the behavior of the exponential function.
When to Use This Method
Logarithmic transformation is useful when a variable appears both in a positive base and in its exponent, and their separate limits give \(0^0\), \(1^\infty\), or \(\infty^0\). It often turns a difficult-looking power into a product or quotient involving logarithms. The quotient may then be suitable for L'Hôpital's Rule, but it may also be simpler to evaluate by algebra or a known limit.
Keep the logical order clear: first establish the limit of \(g(x)\ln f(x)\); then apply the exponential function. If that transformed limit does not exist, do not assume the original power has a limit. Further analysis may be needed.
Check Your Understanding
For each question, consider the base and exponent separately before applying logarithmic transformation where appropriate.
- For \(x\to0^+\), identify the indeterminate form of \((2x)^x\). What expression should be evaluated after taking logarithms?
- Evaluate \(\displaystyle\lim_{x\to\infty}\left(1+\frac{2}{x}\right)^x\) by taking the logarithm and finding its limit.
- For \(x\to\infty\), what does a transformed limit of \(g(x)\ln f(x)\to-\infty\) imply about \(f(x)^{g(x)}\)?
- Why is positivity of the base required when using the real natural logarithm in this method?
- Explain why a base tending to \(1\) does not, by itself, determine the limit of a power whose exponent tends to infinity.
