Why Mixed Limits Need a Plan
A limit expression can contain several familiar pieces and still be difficult to evaluate if they are handled in the wrong order. A polynomial factor may approach zero at the same time as a trigonometric factor, so direct substitution can produce a quotient of zero by zero. The useful strategy is to simplify the algebra and rewrite the trigonometric part into a form whose limit is already known.
For example, the standard sine limit applies to a quotient of the form \(\sin u/u\) as \(u\) tends to zero. A factor such as \(\sin(5x)/x\) is not quite in that form, but it can be rewritten as a constant times \(\sin(5x)/(5x)\). Factoring polynomial expressions and making this kind of rewrite lets us apply limit laws in a justified sequence.
Facts Used in Mixed Calculations
Throughout, trigonometric angles are measured in radians. The standard sine limit is
Two useful consequences are the scaled sine limit and its reciprocal form. If \(c\ne 0\), then
For the first formula, write \(\sin(cx)/x=c\sin(cx)/(cx)\), and use the standard sine limit as \(cx\to 0\). For the reciprocal formula, the first limit is nonzero, so taking reciprocals is valid near the limit point.
We will also use the standard cosine facts \(\lim_{x\to 0}\cos x=1\) and \(\lim_{x\to 0}(1-\cos x)/x^2=1/2\). The second is the quadratic cosine limit. Since \(\tan u=\sin u/\cos u\) wherever \(\cos u\ne 0\), these facts also give \(\lim_{u\to 0}\tan u/u=1\).
A Reliable Sequence of Steps
In a mixed calculation, the key is to make each transformation exact for inputs near the limit point, excluding the point itself if the original expression is undefined there. If two expressions are equal for all such nearby inputs, they have the same limit, provided the limit exists. This allows us to cancel a factor without claiming that the original expression is defined at the point.
Check whether the expression has an ordinary finite value or instead gives a form such as zero over zero.
Factor polynomial expressions and cancel only factors that are nonzero in the punctured neighborhood being considered.
Use substitutions such as \(u=cx\) to produce \(\sin u/u\), or use an established cosine or tangent limit.
Check that any denominator limit used in a quotient law is nonzero, and confirm that the rewritten expression agrees with the original nearby.
Worked Examples
Worked Example: A Polynomial Factor and a Scaled Sine Quotient
Evaluate \(\displaystyle \lim_{x\to 0}\frac{x^2+3x}{x}\frac{\sin(5x)}{10x}\).
Direct substitution makes the first quotient \(0/0\), so do not apply the quotient law to it. Factor the numerator:
For \(x\ne 0\), cancel the common factor \(x\). Also rewrite the trigonometric quotient to use the standard sine limit:
As \(x\to 0\), the first factor tends to \(3/2\), and the sine quotient tends to \(1\). Therefore, by the product law,
Check: The displayed rewrite is an equality for \(x\ne 0\), so it preserves the limit. The normalized sine quotient tends to \(1\), leaving the factor \((x+3)/2\), whose limit is \(3/2\).
Worked Example: Factoring Near a Cosine Zero
Evaluate \(\displaystyle \lim_{x\to \pi/2}\frac{x^2-\pi^2/4}{\cos x}\).
At \(x=\pi/2\), both numerator and denominator are zero. Set \(h=x-\pi/2\), so \(h\to 0\) as \(x\to\pi/2\). Factor the difference of squares and use \(\cos(\pi/2+h)=-\sin h\):
The standard sine limit gives \(\sin h/h\to 1\), and therefore \(h/\sin h\to 1\). The other factor tends to \(\pi\). Thus
Check: For \(h\ne 0\) near zero, the original quotient equals \(-(x+\pi/2)h/\sin h\). Its factors tend to \(-\pi\) and \(1\), respectively, confirming the result.
Worked Example: A Quadratic Cosine Limit and a Sine Quotient
Evaluate \(\displaystyle \lim_{x\to 0}\frac{1-\cos(4x)}{x\sin(2x)}\).
Substitution gives zero in both numerator and denominator. Separate the expression into factors whose limits are known. For \(x\ne 0\) near zero,
In the first factor, \(4x\to 0\), so the quadratic cosine limit gives \(1/2\). In the second factor, \(\sin(2x)/(2x)\to 1\), so its reciprocal \(2x/\sin(2x)\) also tends to \(1\). The product law now applies:
Check: The two factors on the right multiply to the original expression for \(x\ne 0\). Their limits are \(1/2\) and \(8\), so their product is \(4\).
Worked Example: Combining a Polynomial with Tangent
Evaluate \(\displaystyle \lim_{x\to 0}\frac{x^2-4x}{\tan(3x)}\).
Factor the numerator and write tangent as sine divided by cosine. For \(x\ne 0\) sufficiently close to zero,
The factors tend to \(-4\), \(1\), and \(1/3\), respectively. The last limit follows from the reciprocal form of the scaled sine limit. Applying the product law gives
Check: The rewritten expression equals the original whenever the quotients are defined near, but not at, zero. Its factor limits multiply to \(-4/3\), as stated.
Why the Order of Operations Matters
The examples use different arrangements of the same basic ideas. In one, factoring exposes a canceling polynomial factor. In another, shifting the variable turns a cosine near \(\pi/2\) into a sine near zero. In the remaining examples, a rescaling puts the trigonometric quotient into standard form. None of these steps changes the value of the limit because the rewritten expressions agree with the originals for nearby inputs where they are defined.
A common mistake is to treat \(0/0\) as though it were an ordinary number. For example, direct substitution into the first example might be written incorrectly as
This calculation is invalid: \(0/0\) is undefined, not zero. It also does not say whether the limit exists or what its value is. The correction is to factor and rewrite for \(x\ne 0\), then evaluate the limit of the resulting expression. In that example, the valid calculation gives \(3/2\).
Check Your Understanding
For each question, identify the useful rewrite before applying limit laws.
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{x^2+2x}{x}\frac{\sin(3x)}{6x}\).
- Evaluate \(\displaystyle \lim_{x\to 0}\frac{x^2-9x}{\sin(3x)}\).
- Why is the quotient law not enough to evaluate a limit when both the numerator and denominator tend to zero?
- In \(\displaystyle \lim_{x\to 0}\frac{1-\cos(2x)}{x\sin x}\), which standard limits can be used after rewriting the expression as a product?
- State the condition on \(c\) needed to use \(\displaystyle \lim_{x\to 0}\sin(cx)/x=c\).
