Why a Zero-to-Zero Power Needs a Method
A power whose base tends to zero and whose exponent also tends to zero does not have a limit determined by those two facts alone. The base becomes small, but an exponent close to zero can weaken that decrease substantially. The result depends on how quickly the base and exponent change relative to one another.
For instance, as \(x\to 0^+\), both \(x\) and \(x^2\) tend to zero, and both \(x\) and \(1/\ln x\) tend to zero. But the associated powers need not behave alike. We need a way to compare the exponent with the logarithm of the base.
The positivity condition matters: the natural logarithm \(\ln A(x)\) is defined as a real number only when \(A(x)>0\). In this tutorial, assume the base is positive throughout a sufficiently small punctured neighborhood of the point under consideration. The exponent may be positive or negative, provided it tends to zero.
The Logarithm Converts the Power into a Product
Let \(A(x)>0\) and let \(B(x)\) be real. The logarithm and exponential identities give
Because \(A(x)\to0^+\), its logarithm satisfies \(\ln A(x)\to-\infty\). Meanwhile, \(B(x)\to0\). Thus the exponent in the exponential, \(B(x)\ln A(x)\), has the indeterminate product form \(0\cdot(-\infty)\). Evaluating this product is the central task.
The finite-limit case can give any positive finite answer, not just \(1\). The other two cases show why it is not safe to guess the result from the form \(0^0\) alone.
Why the Logarithm Method Works
Set \(Q(x)=B(x)\ln A(x)\). The power is exactly \(e^{Q(x)}\) wherever the base is positive. If \(Q(x)\to L\) for finite \(L\), continuity of the exponential function gives \(e^{Q(x)}\to e^L\). If \(Q(x)\to-\infty\), then for every positive \(\varepsilon\), eventually \(Q(x)<\ln\varepsilon\), so \(e^{Q(x)}<\varepsilon\). Since \(e^{Q(x)}>0\), this proves \(e^{Q(x)}\to0\). If \(Q(x)\to+\infty\), then for every real \(M>0\), eventually \(Q(x)>\ln M\), so \(e^{Q(x)}>M\). Hence \(e^{Q(x)}\to+\infty\).
This also explains the method's logic: it does not approximate the power by a different expression. It uses an exact identity, evaluates the exponent of the exponential, and then applies the behavior of the exponential function.
Worked Examples
Worked Example: A power tending to one
Evaluate \(\displaystyle\lim_{x\to0^+}x^x\).
Here the base \(x\to0^+\), and the exponent \(x\to0\). Take the logarithmic product:
To evaluate it using L’Hôpital’s rule, write it as a quotient:
As \(x\to0^+\), both numerator and denominator tend to \(-\infty\) and \(+\infty\), respectively, so this is an \(\infty/\infty\) form. Both functions are differentiable for \(x>0\), and the derivative of the denominator, \(-1/x^2\), is nonzero there. The derivative quotient is
Therefore \(x\ln x\to0\), and the logarithm method gives
Check: the logarithm of the original power is \(x\ln x\), whose limit is \(0\); exponentiating that limit gives \(1\), as obtained.
Worked Example: A finite limit other than one
Evaluate \(\displaystyle\lim_{x\to0^+}(x^2)^{3/\ln x}\).
The base \(x^2\) is positive and tends to zero. Since \(\ln x\to-\infty\), the exponent \(3/\ln x\to0\). The logarithmic product is
Thus the product is exactly \(6\) for every \(x>0\), and the power has limit
Check: taking the logarithm of the power gives \(6\), so its value is \(e^6\); this is positive and finite, as the theorem predicts for a finite logarithmic-product limit.
Worked Example: A power tending to zero
Evaluate \(\displaystyle\lim_{x\to0^+}x^{1/\sqrt{-\ln x}}\).
For \(0<x<1\), the quantity \(-\ln x\) is positive. As \(x\to0^+\), it tends to \(+\infty\), so \(1/\sqrt{-\ln x}\to0\). The logarithmic product is
Therefore
Check: the exponent in the exponential tends to \(-\infty\), so the exponential tends to zero. The original power is positive for every \(x>0\), consistent with a limit of zero.
Worked Example: A power tending to infinity
Evaluate \(\displaystyle\lim_{x\to0^+}x^{-1/\sqrt{-\ln x}}\).
The base still tends to zero, and the exponent \(-1/\sqrt{-\ln x}\) tends to zero. This time the logarithmic product has the opposite sign:
Consequently,
Check: the logarithm of the power tends to \(+\infty\), so the power itself grows without bound. A negative exponent, even one tending to zero, can therefore produce an unbounded result in this form.
When L’Hôpital’s Rule Is Useful
The logarithmic product may simplify directly, as in the examples above. If it does not, it may sometimes be rewritten as a quotient so that L’Hôpital’s rule applies. For example, \(B(x)\ln A(x)\) can be written as \(\ln A(x)/(1/B(x))\) when \(B(x)\ne0\) nearby. Before differentiating, check that the quotient is genuinely an indeterminate \(0/0\) or \(\infty/\infty\) form and that the hypotheses of L’Hôpital’s rule hold on the relevant interval. The conversion is an option, not an automatic instruction to differentiate.
The product may also be easier to evaluate by algebra or a substitution. The examples using \(\sqrt{-\ln x}\) needed no derivative calculations: simplifying the logarithmic product was enough. Choose the simplest valid way to find its limit.
Confirm that the base is positive and tends to zero, while the exponent tends to zero.
Form \(B(x)\ln A(x)\), using \(\ln(A(x)^{B(x)})=B(x)\ln A(x)\).
Simplify, substitute a useful variable, or use L’Hôpital’s rule only when its conditions are met.
A finite limit \(L\) gives \(e^L\); a limit of \(-\infty\) gives zero; a limit of \(+\infty\) gives \(+\infty\).
A Common Mistake
A tempting but incorrect shortcut is to treat \(0^0\) as \(1\), as though the limiting base and exponent could simply be substituted into the expression. Consider the concrete example
The incorrect calculation replaces the base by \(0\), the exponent by \(0\), and declares the answer to be \(0^0=1\). But this does not evaluate the limit: the expression \(0^0\) does not prescribe a limit value. The correct calculation is
so the power tends to \(e^{-\infty}=0\), not \(1\). The decisive information is the limit of the logarithmic product, not the symbols \(0^0\).
Why This Method Matters
Taking logarithms turns a variable power into a product involving the exponent and the logarithm of the base. That product makes the competition between the two changing quantities visible. A base approaching zero does not force the power to approach zero: the power may approach zero, a positive finite number, or infinity, depending on the logarithmic product's limit.
For a finite product limit, remember to exponentiate the answer. If the product tends to \(L\), the power tends to \(e^L\), not \(L\). For an infinite product limit, retain its sign: \(-\infty\) leads to zero, whereas \(+\infty\) leads to infinity.
Check Your Understanding
For each question, begin by identifying the logarithmic product before deciding what happens to the power.
- For \(x\to0^+\), what expression should be evaluated to study \(x^{2x}\)?
- If \(A(x)>0\), \(A(x)\to0\), \(B(x)\to0\), and \(B(x)\ln A(x)\to-3\), what is the limit of \(A(x)^{B(x)}\)?
- What limit does the theorem give if \(B(x)\ln A(x)\to+\infty\)?
- Why is it necessary for the base to be positive when using the real logarithm method?
- For \(x\to0^+\), evaluate the logarithmic product for \(x^{-2/\sqrt{-\ln x}}\), and use it to determine the power's limit.
