Why Some Limits Need More Than One Step
A limit can be difficult not because one calculation is especially complicated, but because its original form does not permit the method you want to use. A product such as \(0\cdot\infty\) is not a quotient, so L’Hôpital’s Rule cannot be applied to it directly. A variable power may have an indeterminate form whose behavior becomes clearer only after taking a logarithm.
In these situations, the main task is to transform the expression without changing its value, then check that the transformed limit has a form to which a known result applies. Sometimes a second transformation is needed after the first one.
When L’Hôpital’s Rule Applies
L’Hôpital’s Rule concerns a quotient, not an arbitrary expression. In its usual one-sided form, suppose \(f\) and \(g\) are differentiable on a punctured interval next to \(a\), \(g(x)\ne 0\) and \(g'(x)\ne 0\) there, and both \(f(x)\) and \(g(x)\) tend to \(0\), or both tend in magnitude to infinity, as \(x\to a\). If the limit of \(f'(x)/g'(x)\) exists, either as a finite value or as an infinite limit, then the limit of \(f(x)/g(x)\) equals that derivative-ratio limit. Corresponding one-sided or infinite-endpoint versions have the same kind of hypotheses.
The essential starting forms are therefore \(0/0\) and \(\infty/\infty\). A product \(0\cdot\infty\), a difference \(\infty-\infty\), or a variable power is not itself one of those forms. Such an expression may sometimes be transformed into a quotient or a logarithmic limit that can be evaluated.
Converting a Product into a Quotient
Suppose \(f(x)\to 0\) and \(|g(x)|\to\infty\). The product \(f(x)g(x)\) has the indeterminate form \(0\cdot\infty\). There are two quotient rewrites, and they do not produce the same form.
With the first rewrite, the numerator tends to \(0\) and the denominator \(1/g(x)\) also tends to \(0\). Thus this quotient has the form \(0/0\), where L’Hôpital’s Rule may apply if its other hypotheses hold. Alternatively, when \(f(x)\ne0\) near the limit point, the second rewrite has numerator and denominator both tending in magnitude to infinity, so it has the form \(\infty/\infty\). In either case, the derivative ratio must still be checked; a rewrite by itself does not establish that the rule applies.
Worked Example: A Product of a Vanishing Factor and a Diverging Factor
Evaluate \(\displaystyle\lim_{x\to0^+}x\ln x\). Here \(x\to0^+\) and \(\ln x\to-\infty\), so the product has the indeterminate form \(0\cdot(-\infty)\).
Use the second quotient rewrite, with \(f(x)=x\) and \(g(x)=\ln x\). Since \(x\ne0\) for \(x>0\), the rewrite is valid:
As \(x\to0^+\), the numerator \(\ln x\to-\infty\) and the denominator \(1/x\to+\infty\), giving the \(\infty/\infty\) type of form required for L’Hôpital’s Rule. The denominator is nonzero, and its derivative \(-1/x^2\) is nonzero for \(x>0\). Differentiate the numerator and denominator:
The derivative ratio is
L’Hôpital’s Rule therefore gives \(\displaystyle\lim_{x\to0^+}\frac{\ln x}{1/x}=0\). Since this quotient equals the original product for every \(x>0\), the requested limit is \(0\). Checking the transformation back, \(x\ln x=\ln x/(1/x)\), confirms that the evaluated quotient is exactly the original expression.
Taking Logarithms of a Variable Power
For a positive variable expression \(A(x)\), a useful identity is \(A(x)^{B(x)}=\exp(B(x)\ln A(x))\). This converts a variable power into an exponential whose exponent is a product. To evaluate a limit of positive powers, one can first find the limit of the logarithm of the expression and then use continuity of the exponential function.
Worked Example: A One-to-Infinity Power
Evaluate \(\displaystyle\lim_{x\to0}(1+x)^{1/x}\). For \(x\) near zero with \(x\ne0\), the base is positive. The base tends to \(1\), while the exponent \(1/x\) is unbounded, so direct substitution gives the indeterminate form \(1^\infty\).
Let \(y=(1+x)^{1/x}\). Taking logarithms gives
As \(x\to0\), the numerator \(\ln(1+x)\to0\) and the denominator \(x\to0\). The quotient is \(0/0\). Both functions are differentiable near zero, and the derivative of the denominator is \(1\), which is nonzero. L’Hôpital’s Rule gives
Thus \(\ln y\to1\). Because \(y=\exp(\ln y)\), continuity of the exponential function gives \(y\to e^1=e\). The result is therefore \(\displaystyle\lim_{x\to0}(1+x)^{1/x}=e\). As a check, taking the logarithm of the proposed limit gives \(\ln e=1\), matching the limit obtained for \(\ln y\).
Chaining Two Transformations
Sometimes a logarithm converts a variable power into an expression that still needs another transformation. The next example chains these steps: first rewrite the power using an exponential, then factor the resulting difference into two quotients with limits that can be evaluated separately.
Worked Example: A Variable Power Inside a Difference Quotient
Evaluate \(\displaystyle\lim_{x\to0}\frac{(1+x)^{1/x}-e}{x}\). Direct substitution into the numerator is not possible because the power has an indeterminate form. The preceding example established the limit of the power, but this quotient also asks how its difference from \(e\) compares with \(x\). A logarithmic transformation will expose the needed rate.
For \(x\ne0\) near zero, set
The logarithmic quotient tends to \(1\), so \(q(x)\to0\). Moreover, \(q(x)\ne0\) for \(x\ne0\) sufficiently near zero: the inequality \(\ln(1+x)<x\), valid for \(x>-1\) with \(x\ne0\), shows that its numerator is nonzero. Rewriting the power as an exponential gives
Consequently, the original quotient can be factored exactly as
For the first factor, \(q(x)\to0\), so the quotient \((e^{q(x)}-1)/q(x)\) has the form \(0/0\). Applying L’Hôpital’s Rule to the function of \(q\) gives
For the second factor, use the definition of \(q(x)\):
The numerator and denominator both tend to zero. Their derivatives are \(1/(1+x)-1\) and \(2x\), respectively, and \(2x\ne0\) when \(x\ne0\). L’Hôpital’s Rule gives
Multiplying the limits of the two factors and the constant \(e\), we obtain
Checking the factorization confirms that no approximation was used: for each permitted \(x\), the product of the two quotients, multiplied by \(e\), equals the original quotient. The evaluated limits of those factors therefore give the limit of the original expression.
Why the Order of Steps Matters
Each transformation above has a specific purpose. Rewriting a product produces a quotient to which L’Hôpital’s Rule may apply. Taking logarithms turns a positive variable power into a limit involving its exponent. Factoring a quotient into simpler factors can reveal separate limits. In every case, the transformation preserves the original expression on the punctured interval under consideration.
It is also important not to apply L’Hôpital’s Rule simply because a limit looks difficult. For example, the rule does not apply directly to \(x\ln x\): it is a product, not a quotient of the required form. Differentiating its two factors separately and adding the results is not a valid use of the rule.
A Reliable Sequence of Checks
When a limit seems to require successive transformations, keep track of both the expression and its form at every stage.
Substitute the limiting behavior of the component parts. Distinguish a determinate form from an indeterminate one.
For a product, consider quotient forms. For a positive variable power, take its logarithm and evaluate the resulting exponent limit.
Before using L’Hôpital’s Rule, confirm that the quotient is \(0/0\) or \(\infty/\infty\), that the functions are differentiable on the relevant punctured interval, and that the denominator derivative is nonzero there.
Find the derivative-ratio limit or the logarithmic limit, then use the exact identity and continuity needed to recover the requested limit.
Check Your Understanding
For each question, identify the form at each stage before deciding which method is valid.
- For \(x\to0^+\), rewrite \(x\ln x\) as a quotient in both ways shown in this tutorial. What form does each quotient have?
- Why is it invalid to apply L’Hôpital’s Rule directly to a product with form \(0\cdot\infty\)?
- In evaluating \(\lim_{x\to0}(1+x)^{1/x}\), which quotient has the form \(0/0\), and what limit must be recovered after evaluating it?
- In the final worked example, what are the limits of the two factors \((e^{q(x)}-1)/q(x)\) and \(q(x)/x\)?
- State two hypotheses, in addition to having a quotient of form \(0/0\) or \(\infty/\infty\), that must be checked when applying L’Hôpital’s Rule.
