Why Infinity-to-Zero Powers Need a Method
A power with a base tending to infinity and an exponent tending to zero has the form \([f(x)]^{g(x)}\), where \(f(x)>0\), \(f(x)\to\infty\), and \(g(x)\to 0\). It is tempting to expect the base’s growth to dominate. But the exponent can approach zero at different rates, so the power may approach a finite number, zero, or infinity.
This is the indeterminate form \(\infty^0\). “Indeterminate” does not mean that the limit does not exist. It means that knowing only the separate limits of the base and exponent is not enough to determine the limit of the power.
Worked Example: A Power Approaching a Finite Value
Evaluate \(\displaystyle \lim_{x\to\infty}x^{2/\ln(x+1)}\). For \(x>0\), the base is positive. Also, \(\ln(x+1)\to\infty\), so \(2/\ln(x+1)\to 0\). This is an infinity-to-zero power.
Set \(y=x^{2/\ln(x+1)}\). Taking the natural logarithm gives
Both logarithms in the quotient tend to infinity. Apply L’Hôpital’s Rule: the functions are differentiable for \(x>0\), and the derivative of the denominator, \(1/(x+1)\), is nonzero there. Thus
Therefore \(\ln y\to 2\cdot 1=2\). Since the exponential function is continuous, \(y=e^{\ln y}\to e^2\). Checking the logarithmic result, \(\ln(e^2)=2\), which is exactly the limit found for \(\ln y\). Hence the original power tends to \(e^2\).
The Logarithm Converts the Power to a Product
The key step is to take the natural logarithm. It is valid because the base is positive, and therefore the real power \(y=[f(x)]^{g(x)}\) is positive. The logarithm rule \(\ln(a^b)=b\ln a\), for \(a>0\), gives an exact identity:
Now the problem is to evaluate \(g(x)\ln(f(x))\). The exponent tends to zero, while \(\ln(f(x))\to\infty\), so this product has the form \(0\cdot\infty\), which is also indeterminate. If the product can be expressed as a quotient with numerator and denominator both tending to infinity in magnitude, L’Hôpital’s Rule may help.
When \(g(x)\ne 0\), the product can be rewritten as
If \(g(x)\to0\) while remaining nonzero near the limiting point, then \(\lvert 1/g(x)\rvert\to\infty\). Since \(\ln(f(x))\to\infty\), the quotient is an infinity-over-infinity form in magnitude. L’Hôpital’s Rule can be used if the numerator and denominator are differentiable near the endpoint, the denominator’s derivative is nonzero there, and the quotient of derivatives has a limit (finite or infinite). The rule does not guarantee that this quotient of derivatives has a limit; it is a method to use when the required limit can be established.
Once the logarithmic limit is known, the result for the power follows from \(y=e^{\ln y}\). In particular, a finite logarithmic limit \(L\) gives the finite power limit \(e^L\). A logarithm tending to \(-\infty\) gives a power tending to \(0\), and one tending to \(+\infty\) gives a power tending to \(+\infty\). These are consequences of the behavior of the exponential function.
Derivation of the Logarithmic Method
The method works because taking the logarithm exactly converts the exponent into a factor. No approximation is involved in this conversion. The limiting step uses continuity of the exponential function, or its behavior at positive and negative infinity.
For each \(x\) near the endpoint, set \(y(x)=[f(x)]^{g(x)}\), with \(f(x)>0\). Then \(y(x)>0\), so \(\ln y(x)\) is defined.
The power rule for logarithms gives \(\ln y(x)=g(x)\ln(f(x))\). Evaluate the limit of this product, using a quotient conversion and L’Hôpital’s Rule only when their conditions hold.
The identity \(y(x)=e^{\ln y(x)}\) holds for every \(x\). If \(\ln y(x)\to L\in\mathbb R\), continuity gives \(y(x)\to e^L\). If \(\ln y(x)\to-\infty\) or \(+\infty\), the corresponding exponential limit is \(0\) or \(+\infty\).
This also explains why the original form is indeterminate: the separate limits \(f(x)\to\infty\) and \(g(x)\to0\) do not specify the limit of their logarithmic product. That product may have different limits for different rates of growth and decay.
When the Logarithmic Product Tends to Zero
Sometimes the exponent goes to zero quickly enough that \(g(x)\ln(f(x))\to0\). In that case the original power tends to \(e^0=1\). The base still tends to infinity; the limit of the power is determined by the interaction between the exponent and the logarithm of the base.
Worked Example: A Power Approaching One
Evaluate \(\displaystyle \lim_{x\to\infty}(x^2+1)^{1/x}\). The base \(x^2+1\) tends to infinity, and \(1/x\to0\). Set \(y=(x^2+1)^{1/x}\). Then
The numerator and denominator both tend to infinity. For \(x>0\), both are differentiable, and the denominator derivative is \(1\ne0\). By L’Hôpital’s Rule,
For the last limit, divide numerator and denominator by \(x^2\): \(2x/(x^2+1)=(2/x)/(1+1/x^2)\to0/1=0\). Hence \(\ln y\to0\), and continuity of the exponential function gives \(y\to e^0=1\). Checking, the logarithm of the proposed limit \(1\) is \(\ln 1=0\), agreeing with the limit of \(\ln y\).
When the Logarithmic Product Is Unbounded
The logarithmic product can also diverge. If it tends to \(+\infty\), then the power grows without bound; if it tends to \(-\infty\), the power shrinks to zero. In both cases the exponent itself can still tend to zero. Its sign and rate of approach matter.
Worked Example: A Power Approaching Zero
Evaluate \(\displaystyle \lim_{x\to\infty}x^{-1/\sqrt{\ln x}}\), considering \(x>1\) so that the exponent is defined. The base tends to infinity, and \(-1/\sqrt{\ln x}\to0\). Put \(y=x^{-1/\sqrt{\ln x}}\). Taking logarithms gives
As \(x\to\infty\), \(\ln x\to\infty\), and therefore \(-\sqrt{\ln x}\to-\infty\). It follows that \(y=e^{\ln y}\to0\). To check the conclusion, for every \(x>1\) the power can also be written \(e^{-\sqrt{\ln x}}\); its exponent tends to \(-\infty\), so the power tends to zero, as found by the logarithmic calculation.
Using L’Hôpital’s Rule Carefully
The conversion \(g(x)\ln(f(x))=\ln(f(x))/(1/g(x))\) is useful, but it has conditions. It requires \(g(x)\ne0\) wherever the quotient is used. For an application of L’Hôpital’s Rule, the quotient must be an appropriate indeterminate form, the functions must be differentiable on a suitable punctured interval, and the denominator’s derivative must not vanish there. After differentiating, check the new quotient and its limit; the original form alone does not guarantee a result.
If \(g(x)\) is zero at points arbitrarily close to the endpoint, the quotient \(1/g(x)\) is not defined at those points. Do not silently apply L’Hôpital’s Rule across them. The identity for the product \(g(x)\ln(f(x))\) remains valid, so one can instead estimate or evaluate that product directly, or analyze intervals on which \(g(x)\ne0\) and separately account for any zero values.
Worked Example: A Power Approaching Infinity
Evaluate \(\displaystyle \lim_{x\to\infty}x^{1/\sqrt{\ln x}}\), for \(x>1\). The base tends to infinity and the exponent tends to zero. Set \(y=x^{1/\sqrt{\ln x}}\). Then
Thus \(y=e^{\ln y}\to+\infty\). To verify the calculation, the identity \(x^{1/\sqrt{\ln x}}=e^{\sqrt{\ln x}}\) follows by taking the exponential of its logarithm; because \(\sqrt{\ln x}\to+\infty\), this expression indeed tends to \(+\infty\).
A Common Mistake: Guessing from the Base or Exponent
A common error is to treat an exponent tending to zero as if the power must tend to \(1\). That conclusion is not justified: it overlooks the factor \(\ln(f(x))\). For example, the exponent in the preceding example tends to zero, but its logarithmic product tends to \(+\infty\), so the power tends to infinity rather than \(1\).
Another mistake is to differentiate the original power directly as a way to evaluate its limit, or to apply L’Hôpital’s Rule to \(g(x)\ln(f(x))\) without first making it a quotient. L’Hôpital’s Rule applies to suitable quotients, not directly to a product. The correction is to take the logarithm, rewrite the product as a quotient only when valid, verify the rule’s conditions, and then recover the power from the logarithmic limit.
Check Your Understanding
For each question, focus first on the limit of the logarithm of the power.
- For \(f(x)>0\), what expression equals \(\ln([f(x)]^{g(x)})\)?
- If \(g(x)\ln(f(x))\to 3\), what is the limit of \([f(x)]^{g(x)}\), and which property justifies the conclusion?
- Why is an infinity-to-zero power form called indeterminate?
- When can the product \(g(x)\ln(f(x))\) be rewritten as a quotient involving \(1/g(x)\)?
- If \(\ln([f(x)]^{g(x)})\to-\infty\), what happens to the original positive power?
