Why vanishing inputs need special care
A sequence can approach zero while its terms remain nonzero. If we substitute a term \(x_n\) into \((e^x-1)/x\) or \(\log(1+x)/x\), both the numerator and denominator approach zero. Direct substitution therefore does not give the limit. The important question is how quickly each numerator approaches zero compared with its denominator.
In these expressions, \(\log\) means the natural logarithm, the inverse function of the exponential. We will establish that both quotients approach \(1\) as their nonzero inputs approach zero. The results apply whether the inputs are positive, negative, or alternate in sign, provided the logarithm is defined.
The exponential difference quotient
We first establish bounds valid on both sides of zero. The basic inequality \(e^t\geq 1+t\) holds for every real \(t\). One justification uses \(g(t)=e^t-1-t\): its derivative is \(g'(t)=e^t-1\), which is negative for \(t<0\) and positive for \(t>0\). Thus \(g\) has its minimum at \(t=0\), where \(g(0)=0\).
For \(x>0\), apply the inequality first with \(t=x\), and then with \(t=-x\). The first application gives \(e^x-1\geq x\). The second gives \(e^{-x}\geq 1-x\); multiplying by \(e^x>0\) gives \(1\geq (1-x)e^x\), or \(e^x-1\leq xe^x\). Dividing by \(x>0\) yields the bounds below.
For \(x<0\), the same two inequalities give bounds in the opposite order after division by \(x\). From \(e^x\geq 1+x\), division by the negative number \(x\) gives \((e^x-1)/x\leq 1\). Also, \(e^{-x}\geq 1-x\) implies \(1\geq (1-x)e^x\), so \(e^x-1\leq xe^x\). Dividing this last inequality by \(x<0\) reverses its direction and gives \((e^x-1)/x\geq e^x\). Therefore:
Both bounds squeeze the quotient toward \(1\) when \(x\to 0\), since \(e^x\to 1\) by continuity. This argument covers inputs of either sign, rather than assuming the sequence approaches zero from just one side.
Why the bounds prove the sequence limit
Here is the full squeeze argument. Each \(x_n\) is either positive or negative. In the positive case, the quotient lies between \(1\) and \(e^{x_n}\); in the negative case, it lies between \(e^{x_n}\) and \(1\). In either case its distance from \(1\) is no greater than the distance from \(e^{x_n}\) to \(1\). Thus \(\left|(e^{x_n}-1)/x_n-1\right|\leq |e^{x_n}-1|\). Since \(x_n\to0\) and the exponential is continuous, \(|e^{x_n}-1|\to0\). The squeeze theorem for sequences now gives the claimed limit.
Worked Example: An alternating exponential input
Evaluate \(\displaystyle\lim_{n\to\infty}\frac{e^{(-1)^n/n}-1}{(-1)^n/n}\).
Set \(x_n=(-1)^n/n\). Since \(|x_n|=1/n\to0\), we have \(x_n\to0\), and \(x_n\neq0\) for every positive integer \(n\). The inputs alternate in sign, so we use the result that applies to both signs:
Substituting the definition of \(x_n\) gives the required limit, \(1\). To check the sign coverage explicitly, even \(n\) give \(0<x_n=1/n\), with \(1\leq (e^{x_n}-1)/x_n\leq e^{x_n}\); odd \(n\) give \(x_n=-1/n<0\), with \(e^{x_n}\leq (e^{x_n}-1)/x_n\leq1\). In both cases the bounding values approach \(1\), confirming the result.
The logarithmic difference quotient
For the logarithmic quotient, the input must satisfy \(x>-1\), because \(1+x\) must be positive. We also exclude \(x=0\), where the displayed quotient is undefined. The logarithmic limit follows from the exponential limit by using the inverse relationship between the two functions.
Suppose \(x_n>-1\), \(x_n\neq0\), and \(x_n\to0\). Define \(y_n=\log(1+x_n)\). Continuity of \(\log\) at \(1\) gives \(y_n\to\log(1)=0\). Since \(x_n\neq0\), we have \(1+x_n\neq1\), and hence \(y_n\neq0\). The inverse relationship gives \(e^{y_n}=1+x_n\), so \(x_n=e^{y_n}-1\). Consequently, the logarithmic quotient can be expressed as the reciprocal of the exponential quotient at \(y_n\).
The exponential result shows that the denominator in the last expression tends to \(1\). Since \(1\neq0\), taking reciprocals preserves the limit, and the logarithmic quotient also tends to \(1\).
Worked Example: A negative logarithmic input
Evaluate \(\displaystyle\lim_{n\to\infty}\frac{\log(1-1/(n+2))}{-1/(n+2)}\).
Set \(x_n=-1/(n+2)\). For every \(n\geq1\), \(x_n<0\) and \(x_n\neq0\). Also, \(n+2\geq3\), so \(1/(n+2)\leq1/3\) and \(x_n\geq-1/3>-1\). Thus the logarithm is defined. Finally, \(|x_n|=1/(n+2)\to0\). The logarithmic theorem applies and gives the limit \(1\).
For a check using the inverse substitution, let \(y_n=\log(1-1/(n+2))\). Then \(e^{y_n}=1-1/(n+2)\), so \(e^{y_n}-1=-1/(n+2)=x_n\). Therefore the original quotient equals \(y_n/(e^{y_n}-1)\), the reciprocal of \((e^{y_n}-1)/y_n\). Because \(y_n\to0\), this reciprocal tends to \(1\), as obtained above.
Combining the two limits
The two results can be used as factors in a larger sequence expression. The main points are to identify the input of each quotient, verify that it is nonzero and tends to zero, and keep track of any constant factors introduced when rewriting the expression.
Worked Example: Comparing an exponential term with a logarithmic term
Evaluate \(\displaystyle\lim_{n\to\infty}\frac{e^{2/n}-1}{\log(1+1/n)}\).
For every positive integer \(n\), both \(2/n\) and \(1/n\) are positive and nonzero, and both tend to zero. Multiply and divide by the inputs of the respective quotients to obtain an exact factorization:
The first factor tends to \(1\) by the exponential result with input \(2/n\). The second factor is the reciprocal of \(\log(1+1/n)/(1/n)\), which tends to \(1\) by the logarithmic result; its limit is therefore \(1\). The third factor is exactly \(2\), since \((2/n)/(1/n)=2\). The product limit is \(1\cdot1\cdot2=2\).
To check the factorization, the first factor contributes \((e^{2/n}-1)n/2\), the second contributes \((1/n)/\log(1+1/n)\), and the third contributes \(2\). Multiplying cancels \(n/2\) against \(2/n\), leaving exactly \((e^{2/n}-1)/\log(1+1/n)\). Thus no extra constant has been introduced, and the computed limit is \(2\).
Common pitfalls
A frequent error is to substitute the limiting input directly and conclude that either quotient has limit \(0/0\), or that the limit does not exist. The expression at the limiting input is indeed undefined, but a sequence limit concerns the values at nonzero inputs near zero. The bounds and inverse-function argument show how those values behave.
Worked Example: Correcting a sign error
A mistaken argument for \(x<0\) might start from \(e^x\geq1+x\), divide by \(x\), and write \((e^x-1)/x\geq1\). This inequality is incorrect because division by a negative number reverses the inequality. The correct calculation is:
The other bound for negative \(x\) comes from \(e^{-x}\geq1-x\). Multiplying by \(e^x>0\) gives \(1\geq(1-x)e^x\), hence \(e^x-1\leq xe^x\). Dividing by \(x<0\) reverses the inequality, giving \((e^x-1)/x\geq e^x\). Together these give \(e^x\leq(e^x-1)/x\leq1\), which correctly squeezes the quotient toward \(1\).
For logarithms, another mistake is to overlook the domain condition. A sequence approaching zero is not by itself enough: every term used in \(\log(1+x_n)\) must satisfy \(x_n>-1\). In applying the logarithmic theorem, check this condition as well as \(x_n\neq0\) and \(x_n\to0\).
What these limits tell us
Both quotients describe a first-order comparison: near zero, \(e^x-1\) has the same leading size as \(x\), and \(\log(1+x)\) has the same leading size as \(x\). In sequence problems, this means either numerator can often be replaced by its small input when evaluating a limit, provided the quotient is arranged and the hypotheses are checked. The replacement is a conclusion about the limit, not an exact equality at finite \(n\).
The sequence formulation is useful because the inputs need not be \(1/n\), need not be positive, and need not approach zero monotonically. They may be any nonzero sequence tending to zero, with the logarithm’s domain condition imposed where necessary. These facts provide reliable tools for limits involving exponential and logarithmic terms without relying on a particular formula for the input sequence.
Check Your Understanding
For each question, first identify the input sequence and verify the conditions for the relevant limit.
- What conditions on \(x_n\) are required to conclude that \((e^{x_n}-1)/x_n\to1\)?
- Why do the bounds for the exponential quotient have different orders when \(x\) is positive and when \(x\) is negative?
- What additional domain condition is needed to apply the logarithmic quotient limit?
- Evaluate \(\displaystyle\lim_{n\to\infty}\frac{\log(1+3/n)}{3/n}\).
- Evaluate \(\displaystyle\lim_{n\to\infty}\frac{e^{1/n}-1}{\log(1+4/n)}\).
