Why Variable Powers Need a Different Approach
A limit can be difficult to evaluate when both the base and the exponent vary. For instance, as \(x\) tends to infinity, the base \(1+1/x\) tends to \(1\), while the exponent \(x\) tends to infinity. The limit of the base alone does not determine the limit of the entire power.
A useful method is to take the natural logarithm. For a positive base \(u(x)\), the identity \(u(x)^{v(x)}=\exp(v(x)\ln u(x))\) turns the power into an expression involving a product. If that exponent has a finite limit, continuity of the exponential function gives the limit of the original power.
A General Result for Positive Variable Powers
For real-valued exponents, a variable power \(u(x)^{v(x)}\) is defined using the natural logarithm when \(u(x)>0\): it means \(\exp(v(x)\ln u(x))\). Thus, the positivity of the base is essential for this real-valued method.
Proof. By the definition of a positive real power, \(u(x)^{v(x)}=\exp(v(x)\ln u(x))\). The exponential function is continuous at every real number. Therefore, as \(v(x)\ln u(x)\) tends to \(L\), its exponential tends to \(\exp(L)=e^L\). This proves the result.
The theorem reduces a variable-power limit to an ordinary limit of a logarithmic expression. That expression may be a product, a quotient, or an indeterminate form that can be rewritten as a quotient. L'Hôpital's Rule is one possible tool for evaluating it, but its conditions must be checked before using it.
The Standard Limit for Powers of \(1+c/x\)
Let \(c\) and \(d\) be fixed real numbers. For all sufficiently large positive \(x\), the base \(1+c/x\) is positive. The logarithm of the power is \[ \ln\left(\left(1+\frac{c}{x}\right)^{dx}\right) =dx\ln\left(1+\frac{c}{x}\right). \] When \(c\ne 0\), this can be written as \(d\) times the quotient \[ \frac{\ln(1+c/x)}{1/x}. \] As \(x\) tends to infinity, both numerator and denominator tend to zero. For \(x\) sufficiently large, they are differentiable and the derivative of the denominator is nonzero. L'Hôpital's Rule at an infinite endpoint therefore applies: \[ \frac{\frac{d}{dx}\ln(1+c/x)}{\frac{d}{dx}(1/x)} = \frac{-c/(x^2(1+c/x))}{-1/x^2} = \frac{c}{1+c/x} \longrightarrow c. \] So the logarithm of the power tends to \(dc\). The theorem then yields the standard limit below. If \(c=0\), the base is exactly \(1\), and the same conclusion holds directly.
This formula requires \(c\) and \(d\) to be fixed real numbers, with \(x\) tending to positive infinity. The base is positive eventually even when \(c\) is negative: for sufficiently large \(x\), \(|c/x|<1\), so \(1+c/x>0\). The exponent \(d\) may be positive, negative, or zero.
Worked Examples
Worked Example: Evaluating \(x^{1/x}\) at Infinity
Find \(\lim_{x\to\infty}x^{1/x}\). The base \(x\) is positive for \(x>0\), so take the logarithm. The logarithm of the power is \[ \ln(x^{1/x})=\frac{\ln x}{x}. \] As \(x\) tends to infinity, \(\ln x\) and \(x\) both tend to infinity. They are differentiable for \(x>0\), and the derivative of the denominator is \(1\), which is nonzero. L'Hôpital's Rule gives \[ \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = \lim_{x\to\infty}\frac{1}{x} =0. \] Thus the logarithm of the original power tends to \(0\). Exponentiating, \[ \lim_{x\to\infty}x^{1/x}=e^0=1. \]
Check: the logarithm of the proposed limiting value \(1\) is \(\ln 1=0\), which agrees with the limit just found for \(\ln(x^{1/x})\). The positive-power theorem therefore confirms the result.
Worked Example: A Fixed Multiple in the Exponent
Evaluate \[ \lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}. \] For sufficiently large \(x\), the base is positive. Taking logarithms gives \[ \ln\left(\left(1+\frac{3}{x}\right)^{2x}\right) =2x\ln\left(1+\frac{3}{x}\right) =2\frac{\ln(1+3/x)}{1/x}. \] The numerator and denominator in the quotient both tend to zero. The functions are differentiable for sufficiently large \(x\), and the denominator derivative \(-1/x^2\) is nonzero. Therefore L'Hôpital's Rule applies: \[ \begin{aligned} \lim_{x\to\infty}\frac{\ln(1+3/x)}{1/x} &=\lim_{x\to\infty} \frac{-3/(x^2(1+3/x))}{-1/x^2}\\ &=\lim_{x\to\infty}\frac{3}{1+3/x}\\ &=3. \end{aligned} \] Multiplying this logarithmic limit by \(2\), the logarithm of the original power tends to \(6\). Hence \[ \lim_{x\to\infty}\left(1+\frac{3}{x}\right)^{2x}=e^6. \]
Check: the logarithm of the claimed answer \(e^6\) is \(6\), exactly the limit obtained for the logarithm of the original expression.
Worked Example: A Smaller Additional Term in the Exponent
Evaluate \[ \lim_{x\to\infty}\left(1+\frac{2}{x}\right)^{x+\sqrt{x}}. \] The base is positive for \(x>0\). The logarithm of the power is \[ (x+\sqrt{x})\ln\left(1+\frac{2}{x}\right) = x\ln\left(1+\frac{2}{x}\right) + \sqrt{x}\ln\left(1+\frac{2}{x}\right). \] For the first term, write \[ x\ln\left(1+\frac{2}{x}\right) = \frac{\ln(1+2/x)}{1/x}. \] Both parts of this quotient tend to zero. Differentiating numerator and denominator gives \[ \lim_{x\to\infty}\frac{\ln(1+2/x)}{1/x} = \lim_{x\to\infty} \frac{-2/(x^2(1+2/x))}{-1/x^2} = \lim_{x\to\infty}\frac{2}{1+2/x} =2. \] For the second term, write \[ \sqrt{x}\ln\left(1+\frac{2}{x}\right) = \frac{\ln(1+2/x)}{1/\sqrt{x}}. \] Again, numerator and denominator tend to zero. Their derivatives are \(-2/(x^2(1+2/x))\) and \(-1/(2x^{3/2})\), respectively. The derivative quotient is \[ \frac{-2/(x^2(1+2/x))}{-1/(2x^{3/2})} = \frac{4}{\sqrt{x}(1+2/x)} \longrightarrow 0. \] L'Hôpital's Rule applies because these functions are differentiable for \(x>0\) and the denominator derivative is nonzero there. The logarithm of the original power therefore tends to \(2+0=2\). Exponentiating gives \[ \lim_{x\to\infty}\left(1+\frac{2}{x}\right)^{x+\sqrt{x}}=e^2. \]
Check: the two terms in the logarithm tend to \(2\) and \(0\), so their sum tends to \(2\). The logarithm of the proposed answer, \(e^2\), is also \(2\).
Why the Logarithm Step Matters
A common error is to reason that if a base tends to \(1\), then a power with that base must also tend to \(1\). This ignores the fact that the exponent may vary without bound. For example, the incorrect calculation \[ \left(1+\frac{1}{x}\right)^x\longrightarrow 1^x=1 \] treats \(x\) as though it were fixed while taking the limit. But \(x\) is changing, so this substitution does not follow from continuity of a fixed-power function. In fact, taking logarithms shows that \[ x\ln\left(1+\frac{1}{x}\right) = \frac{\ln(1+1/x)}{1/x} \longrightarrow 1, \] and hence the original power tends to \(e\), not \(1\).
Another mistake is to use L'Hôpital's Rule merely because an expression looks difficult. The rule applies to a quotient whose numerator and denominator both tend to zero or both tend in magnitude to infinity, with the required differentiability and nonvanishing denominator derivative nearby. In the examples above, the quotient form and its indeterminate form were stated before differentiating.
A Reliable Procedure
Confirm that the base is positive near the limit, so the real logarithm is defined.
Use \(\ln(u^v)=v\ln u\) for positive \(u\).
Rewrite products as quotients if needed, and verify the conditions before applying L'Hôpital's Rule.
If the logarithmic expression tends to a finite \(L\), the original power tends to \(e^L\).
Confirm that the logarithm of the proposed limit agrees with the limit found for the logarithm of the expression.
When This Method Is Useful
Logarithms are especially useful when a limit has a base approaching \(1\) and an exponent growing in magnitude, or when both parts vary in a way that makes direct substitution inconclusive. The transformed expression \(v(x)\ln u(x)\) often reveals the balance between how quickly the base approaches \(1\) and how quickly the exponent changes.
The method also applies when the transformed logarithmic expression splits into several terms, as in the third example. Each term can be studied separately, provided the limits exist and the sum is handled accordingly. Once a finite limit for the logarithm has been established, exponentiation completes the argument; no separate guess about the original power is needed.
Check Your Understanding
For each question, identify the logarithmic expression that should be evaluated before finding the original power limit.
- Evaluate \(\lim_{x\to\infty}x^{2/x}\).
- For fixed real \(c\) and \(d\), what is \(\lim_{x\to\infty}(1+c/x)^{dx}\), and why is the base positive for all sufficiently large \(x\)?
- For \(\lim_{x\to\infty}(1+1/x)^{3x}\), state the quotient to which L'Hôpital's Rule is applied and the limit of its derivative quotient.
- Why does the fact that \(u(x)\) tends to \(1\) not by itself determine the limit of \(u(x)^{v(x)}\) when \(v(x)\) varies?
- If \(u(x)>0\) and \(v(x)\ln u(x)\) tends to \(L\), what is the limit of \(u(x)^{v(x)}\), and which property justifies the final step?
