Exponential and Logarithmic Limits Near One

Introduction

As \(x\) approaches zero, \(e^{ax}\) approaches \(1\), and \(1+ax\) also approaches \(1\). Consequently, both \(e^{ax}-1\) and \(\ln(1+ax)\) approach zero. Dividing either expression by \(x\) gives a quotient of the indeterminate form \(0/0\).

L’Hôpital’s Rule turns these quotients into straightforward limits of derivatives. The parameter \(a\) matters: it determines the value and sign of the limit, and when \(a=0\), the numerator is identically zero. For the logarithm, we must also check that its argument is positive near the point of approach.

The Two Basic Limits

Throughout, \(\ln\) denotes the natural logarithm, whose derivative is \(1/u\) at a positive input \(u\). The exponential derivative rule is \(\frac{d}{dx}e^{u(x)}=e^{u(x)}u'(x)\). These rules let us differentiate the numerators below.

When \(a=0\), both numerators are zero for every \(x\) in their domains, so both quotients are zero for \(x\ne 0\), and both limits equal zero. When \(a\ne 0\), the quotients have the form \(0/0\) at the limit point. L’Hôpital’s Rule applies if the functions are differentiable on a punctured neighborhood, the denominator derivative is nonzero there, and the limit of the derivative quotient exists.

For these particular quotients, the denominator is \(x\), whose derivative is \(1\), so the denominator derivative does not vanish. The exponential numerator is differentiable for all real \(x\). The logarithmic numerator is differentiable wherever \(1+ax>0\), which includes a sufficiently small neighborhood of zero.

Why the Limits Equal the Parameter

For \(a\ne 0\), first consider the exponential quotient. As \(x\to 0\), continuity of the exponential function gives \(e^{ax}-1\to e^0-1=0\), and \(x\to 0\). Thus the quotient is \(0/0\). Differentiate the numerator and denominator with respect to \(x\): the numerator derivative is \(ae^{ax}\), and the denominator derivative is \(1\). Therefore L’Hôpital’s Rule gives

For the logarithmic quotient, continuity of \(\ln\) at \(1\) gives \(\ln(1+ax)\to\ln(1)=0\), while \(x\to 0\). The chain rule gives the numerator derivative \(\frac{a}{1+ax}\), as long as \(1+ax>0\). Applying L’Hôpital’s Rule yields

For \(a=0\), the result follows directly as noted above, so the formulas hold for every real \(a\). This argument explains why the parameter survives in the answer: differentiating the inner expression \(ax\) contributes a factor of \(a\).

Worked Examples

Parameter Choices and Sign Checks

The formulas also handle zero and negative parameter values without changing their form. For either basic quotient, a positive \(a\) gives a positive limit, a negative \(a\) gives a negative limit, and \(a=0\) gives a zero limit. The sign can also be checked directly: near zero, \(e^{ax}-1\) and \(\ln(1+ax)\) have the sign of \(ax\), while the denominator \(x\) has the sign of \(x\). When \(a\ne 0\), the quotient therefore has the sign of \(a\).

Constant factors in the denominator can be handled after applying the same reasoning. If \(b\ne 0\), then

Indeed, each quotient is \(1/b\) times the corresponding basic quotient. The condition \(b\ne 0\) matters: if \(b=0\), the denominator \(bx\) is identically zero, and the quotient is not defined.

A Common Mistake: Losing the Chain-Rule Factor

A frequent error is to differentiate \(e^{ax}\) as \(e^{ax}\), or \(\ln(1+ax)\) as \(1/(1+ax)\), omitting the derivative of the inner function. For example, an incorrect calculation for the first worked example might be

This is wrong because \(\frac{d}{dx}e^{-3x}=-3e^{-3x}\), not \(e^{-3x}\). The correct calculation is

The same issue occurs with logarithms: \(\frac{d}{dx}\ln(1-2x)=\frac{-2}{1-2x}\), not \(\frac{1}{1-2x}\). Keeping the chain-rule factor is essential because it carries the parameter into the limit.

When to Use This Method

These limits are useful whenever an expression contains an exponential or logarithm approaching its value at one and the overall quotient becomes \(0/0\). Before using L’Hôpital’s Rule, check the form by evaluating the numerator and denominator limits, verify that the functions are differentiable in the required punctured neighborhood, and confirm that the denominator derivative is nonzero there.

Do not apply the rule merely because a quotient contains an exponential or logarithm. The relevant condition is the indeterminate form \(0/0\) or \(\infty/\infty\), together with the rule’s differentiability and derivative conditions. In the examples here, the original numerators and denominators both approach zero, and differentiating them produces a derivative quotient with an ordinary finite limit.