Parameters and Indeterminate Limits
A limit can depend on constants that are fixed while the variable approaches a point. The limit is taken with respect to the variable; the constants are not changing during that process. After evaluating the limit, the answer may be a formula in those constants.
A common example is a quotient of exponential expressions. At \(x=0\), both \(e^{ax}-1\) and \(e^{bx}-1\) equal zero, so their quotient has the indeterminate form \(0/0\), provided it is defined for nearby nonzero \(x\). L’Hôpital’s Rule can then turn the problem into a quotient of derivatives. The rule still requires careful attention to parameter values: in particular, the denominator must not be identically zero near the point.
A Finite-Limit Version of L’Hôpital’s Rule
Here is the version of L’Hôpital’s Rule used in this tutorial. Suppose \(f\) and \(g\) are continuous at \(c\), differentiable on a deleted neighborhood of \(c\), and satisfy \(f(c)=g(c)=0\). Suppose also that \(g(x)\ne 0\) and \(g'(x)\ne 0\) for all \(x\ne c\) sufficiently close to \(c\). If
for a finite real number \(L\), then
The conditions are part of the rule, not optional checks. In particular, it is not enough that both functions have value zero at \(c\): the original quotient must be defined in a deleted neighborhood, and the derivative of the denominator must not vanish there.
Why the Rule Works
The finite-limit conclusion follows from the Cauchy Mean Value Theorem. Fix a nearby \(x\ne c\), and apply that theorem to \(f\) and \(g\) on the interval with endpoints \(c\) and \(x\). The functions are continuous on this closed interval and differentiable in its interior. Thus there is a point \(\xi\) strictly between \(c\) and \(x\) such that
Since \(f(c)=g(c)=0\), this gives
As \(x\to c\), the point \(\xi\), which lies between \(c\) and \(x\), also approaches \(c\). Therefore, if \(f'(t)/g'(t)\to L\) as \(t\to c\), the right-hand side approaches \(L\). The left-hand side consequently approaches \(L\) as well. The requirement that \(g'\) not vanish in the deleted neighborhood ensures that the derivative quotient used in this argument is defined.
The Basic Exponential Family
Consider the family
where \(a\) and \(b\) are fixed real numbers. If \(b\ne 0\), the numerator and denominator both approach zero. Their derivatives are \(a e^{ax}\) and \(b e^{bx}\). Since the exponential function is positive, \(b e^{bx}\ne 0\) for every real \(x\) when \(b\ne0\). Thus, for these parameter values, the derivative quotient is defined near zero and
L’Hôpital’s Rule therefore gives the limit \(a/b\). This classification includes \(a=0\): the numerator is then identically zero, and the limit is zero as long as \(b\ne0\). But \(b=0\) is different. The original denominator is then \(e^0-1=0\) for every \(x\), so the quotient is not defined anywhere. There is no limit of this quotient to evaluate.
Worked Example: A Negative Denominator Parameter
Evaluate
At \(x=0\), the numerator is \(e^0-1=0\), and the denominator is also \(e^0-1=0\). For \(x\ne0\), the denominator is nonzero near zero: \(e^{-2x}=1\) only when \(x=0\). The functions are differentiable, and the denominator derivative \(-2e^{-2x}\) is nonzero for every real \(x\). L’Hôpital’s Rule applies:
As a check, the derivative quotient itself tends to \(-3/2\): its numerator tends to \(3\) and its denominator tends to \(-2\). These are the first-order rates at which the original numerator and denominator approach zero.
When One Application Still Gives 0/0
Some parameterized quotients have numerators and denominators whose first derivatives also vanish at the limit point. In that situation, one application of the rule may leave an indeterminate form. If the resulting derivative quotient satisfies the hypotheses of L’Hôpital’s Rule, the rule can be applied again.
Worked Example: A Quotient Requiring Two Applications
For a fixed \(b\ne0\), evaluate
Both expressions vanish at zero. The original denominator is nonzero for \(x\ne0\): for every nonzero real \(u\), \(e^u>1+u\), so \(e^{bx}-1-bx>0\) when \(x\ne0\). Both functions are differentiable. Their first derivatives are
At zero, both first derivatives equal zero. For \(x\ne0\), the denominator derivative \(b(e^{bx}-1)\) is nonzero because \(b\ne0\) and \(e^{bx}\ne1\). Thus the first application is valid, and it gives
The quotient on the right is still \(0/0\) at zero. Its numerator and denominator are differentiable, and the derivative of its denominator is \(b^2e^{bx}\), which is nonzero for every real \(x\) because \(b\ne0\). Applying L’Hôpital’s Rule a second time gives
The result is parameter-dependent, and the condition \(b\ne0\) is essential. If \(b=0\), the original denominator is identically zero. As a check on the value, the second derivatives of the original numerator and denominator at zero are \(4\) and \(b^2\), respectively; their ratio is \(4/b^2\), matching the computed limit.
Parameters That Cancel the Numerator
A parameter may make the numerator vanish identically, or make its derivative vanish at the limit point. Such cases should be interpreted using the original expression and the rule’s hypotheses, rather than by assuming that every coefficient in the answer must be nonzero.
Worked Example: A Cancellation in the Numerator
Evaluate the limit for fixed real \(a,c\) and fixed \(b\ne0\):
At zero, both numerator and denominator equal zero. The denominator is nonzero for \(x\ne0\), since \(e^{bx}=1\) for real \(x\) only when \(x=0\). Its derivative \(be^{bx}\) is nonzero for every \(x\) because \(b\ne0\). Differentiating numerator and denominator gives
Taking the limit of this derivative quotient yields
Therefore the original limit is \((a-c)/b\). For a numerical instance, take \(a=2\), \(c=-1\), and \(b=4\). Substitution into the parameter formula gives \((2-(-1))/4=3/4\). The derivative quotient confirms this: its numerator tends to \(2-(-1)=3\), and its denominator tends to \(4\). If \(a=c\), the original numerator is identically zero and the formula correctly gives zero, provided \(b\ne0\).
Common Pitfall: Ignoring the Denominator Parameter
A tempting but incorrect calculation is to differentiate the basic exponential quotient and then write \(a/b\) without checking whether \(b\) is zero. For example, setting \(a=2\) and \(b=0\) would produce the meaningless expression \(2/0\). More importantly, the original quotient is
which is undefined for every \(x\). There is no deleted neighborhood on which this quotient exists, and its denominator derivative is identically zero. Both facts rule out applying L’Hôpital’s Rule.
The correction is to check the denominator before differentiating. For \(b\ne0\), the denominator is nonzero for \(x\ne0\) near zero and its derivative is \(be^{bx}\ne0\), so the parameter formula \(a/b\) applies. For \(b=0\), the quotient itself is undefined and no such formula applies.
A Reliable Parameter Check
For a parameterized 0/0 problem, keep the following checks tied to the actual expression. A parameter value that makes a numerator derivative zero does not automatically invalidate the rule. A parameter value that makes the denominator identically zero does.
Identify which symbols are constants and which variable is approaching the limit point.
Confirm that numerator and denominator approach zero, and that the original denominator is nonzero in a deleted neighborhood.
Verify that it does not vanish in that deleted neighborhood before applying L’Hôpital’s Rule.
If it remains \(0/0\), check the hypotheses again before applying the rule a second time.
Include conditions such as \(b\ne0\) alongside the resulting formula.
The value of a parameter can change whether a quotient is defined, how many applications of the rule are needed, or whether the numerator cancels. Treating parameter cases explicitly makes the conclusion precise and prevents a formally plausible calculation from being applied outside its valid conditions.
Check Your Understanding
For each question, identify the parameter restrictions and the relevant hypotheses before using the result.
- For fixed \(a,b\), what is the limit of \((e^{ax}-1)/(e^{bx}-1)\) as \(x\to0\) when \(b\ne0\)? Why is \(b=0\) excluded?
- In the two-application example, what are the first derivatives at zero, and why does the first application still leave an indeterminate form?
- For \((e^{ax}-e^{cx})/(e^{bx}-1)\), what happens when \(a=c\), assuming \(b\ne0\)?
- Why must the derivative of the denominator be nonzero in a deleted neighborhood when applying the finite-limit version of L’Hôpital’s Rule?
- For fixed \(b\ne0\), find the limit of \((e^{5x}-1)/(e^{bx}-1)\) as \(x\to0\), and state which parameter condition makes the formula valid.
