Why Infinity Over Infinity Needs Care
When both the numerator and denominator of a quotient grow without bound, their limits alone do not determine the limit of the quotient. One expression may grow faster than the other, or they may grow at comparable rates. For example, as \(x\to\infty\), the quotient \(x/x\) approaches \(1\), while \(x/x^2\) approaches \(0\) and \(x^2/x\) grows without bound.
L'Hôpital's Rule compares the rates of change of the numerator and denominator. It can be powerful, but only when the original quotient has the required indeterminate form and the differentiability conditions hold.
Eligibility for the Infinity-Over-Infinity Form
Suppose \(x\) approaches an endpoint \(a\), or \(x\to\infty\). The quotient \(f(x)/g(x)\) has the infinity-over-infinity form if both \(f(x)\) and \(g(x)\) tend to positive infinity, both tend to negative infinity, or each tends to an infinite value. In informal notation, its form is \(\infty/\infty\).
The rule also applies to a quotient whose numerator and denominator both tend to negative infinity. The essential form is that both magnitudes grow without bound; the signs of the original functions do not by themselves determine the quotient's limit.
Before using the rule, check three points: the original numerator and denominator both tend to infinity in magnitude; both functions are differentiable near the limit; and the derivative of the denominator is nonzero there. These checks establish eligibility. Once the rule is applied once, evaluate \(f'/g'\) on its own merits. A further application would require its own eligibility check.
Why the Rule Works at Infinity
A short derivation explains why comparing derivatives can determine the original ratio. Assume \(f(x)\to\infty\), \(g(x)\to\infty\), and \(f'(x)/g'(x)\to L\), where \(L\) is finite. Fix a sufficiently large number \(A\). For any \(x>A\), the Cauchy Mean Value Theorem gives a number \(c\) between \(A\) and \(x\) such that
For any positive tolerance \(\varepsilon\), the derivative quotient is within \(\varepsilon\) of \(L\) whenever its input is sufficiently large. Choose \(A\) beyond that threshold. Then for every \(x>A\), the point \(c\) is also beyond the threshold, so the difference quotient on the left is within \(\varepsilon\) of \(L\). Finally, rewrite the original ratio as
As \(x\to\infty\), the second factor in the product approaches \(1\), because \(g(A)\) is fixed while \(g(x)\to\infty\). The final term approaches \(0\). The first factor approaches \(L\) by the Cauchy Mean Value Theorem argument. Therefore \(f(x)/g(x)\to L\). This gives the reason for the rule: over a long interval, the ratio of the total changes is controlled by the ratio of the instantaneous rates of change.
Worked Example: A Rational Quotient
Worked Example: A Rational Quotient
Evaluate \(\lim_{x\to\infty}\frac{3x^2+4x-1}{5x^2-2x+6}\).
First check the form. The leading terms show that the numerator tends to infinity and the denominator tends to infinity, so the quotient is eligible for L'Hôpital's Rule. Both polynomials are differentiable. Their derivatives are
The derivative of the denominator is nonzero for \(x>1/5\), so it is nonzero near \(x=\infty\). Applying the rule once gives
Divide the numerator and denominator of the new quotient by \(x\):
Check the answer by dividing the original numerator and denominator by \(x^2\):
The independent simplification confirms the result.
Worked Example: An Exponential Quotient
Worked Example: An Exponential Quotient
Evaluate \(\lim_{x\to\infty}\frac{2e^x+3x}{5e^x-x}\).
As \(x\to\infty\), both \(e^x\) and \(x\) grow without bound, so the numerator and denominator both tend to infinity. The functions are differentiable, and the derivative of the denominator is \(5e^x-1\), which is positive for sufficiently large \(x\). L'Hôpital's Rule therefore applies:
Divide the numerator and denominator by \(e^x\):
To check the result independently, divide the original quotient by \(e^x\) in both numerator and denominator:
Here \(xe^{-x}\to0\), since the exponential in the denominator grows faster than the linear factor. The check agrees with the L'Hôpital calculation.
Worked Example: A Logarithmic Quotient
Worked Example: A Logarithmic Quotient
Evaluate \(\lim_{x\to\infty}\frac{(\ln x)^2}{\ln(x^2+1)}\).
For \(x>1\), both logarithmic expressions are defined. As \(x\to\infty\), \((\ln x)^2\to\infty\), and \(\ln(x^2+1)\to\infty\). The denominator's derivative is \(2x/(x^2+1)\), which is positive for \(x>0\). The quotient meets the conditions for one application of L'Hôpital's Rule:
Simplify the quotient of derivatives by multiplying by the reciprocal:
Thus the original quotient tends to infinity. As a check, for \(x>0\), \(\ln(x^2+1)=2\ln x+\ln(1+1/x^2)\). The second term tends to \(0\), while \(2\ln x\to\infty\). Therefore the original quotient grows like \((\ln x)^2/(2\ln x)=(\ln x)/2\), which also tends to infinity.
A Common Mistake: Differentiating Only the Numerator
L'Hôpital's Rule uses the quotient of the derivatives, \(f'(x)/g'(x)\). It does not say to differentiate only the numerator while keeping the original denominator. That shortcut can give a different and incorrect result.
For example, consider \(\lim_{x\to\infty}x^2/x\). The original quotient is infinity over infinity, so L'Hôpital's Rule applies. A mistaken calculation might differentiate the numerator but leave the denominator unchanged:
This is not the rule. The correct calculation differentiates both functions:
Direct simplification checks the answer: \(x^2/x=x\), which tends to infinity. The mistaken calculation reached \(2\) because it used neither the original quotient nor the quotient of both derivatives.
What to Check Before Concluding
Confirm that the numerator and denominator each tend to infinity in magnitude.
Confirm differentiability near the limit and that the denominator's derivative is nonzero there.
Form \(f'(x)/g'(x)\), not the derivative of the original quotient.
Compute the derivative quotient's limit, then check it when a direct simplification is available.
L'Hôpital's Rule is valuable because it turns a comparison of unbounded functions into a comparison of their rates of change. The key is to verify eligibility before using it: the same algebraic appearance does not guarantee the same mathematical conditions.
Check Your Understanding
For each question, identify the original form and state the condition that justifies your method.
- For \(\lim_{x\to\infty}\frac{4x^2-x}{7x^2+3}\), what quotient of derivatives results from one application of L'Hôpital's Rule?
- For \(\lim_{x\to\infty}\frac{e^x+x}{3e^x-2x}\), what is the derivative of the denominator, and is it nonzero for all sufficiently large \(x\)?
- For \(\lim_{x\to\infty}\frac{\ln x}{x}\), what is the quotient of derivatives, and what limit does it have?
- Explain why the calculation \(f'(x)/g(x)\) is not the application of L'Hôpital's Rule.
- After applying the rule once, what must be checked before applying it again?
