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Sequence Limits · Tutorial 522 of 998

Mixed Algebraic and Transcendental Limit Problems

Practice separating mixed sequence expressions into factors whose limits are known, while checking when algebraic simplification is needed first.

Beginner 10 min read

What You'll Learn

  • Identify factors in a mixed sequence expression that have standard limits.
  • Use conjugation to simplify a difference of radicals before taking its limit.
  • Rewrite logarithmic, exponential, and sine factors in standard limit forms.
  • Apply product and quotient limit laws only when their conditions hold.
  • Spot and correct the mistake of substituting infinity into an indeterminate expression.

Why Mixed Sequence Limits Need a Plan

A sequence can combine several familiar kinds of expressions: a rational factor, a difference of radicals, a logarithm, an exponential, or a trigonometric factor. The main challenge is often not any one of these pieces. It is choosing an order of operations that reveals their limits without losing important scale information.

A useful strategy is to simplify algebraic expressions first, then rewrite the remaining factors in standard limit forms. Once each factor has a finite limit, the product and quotient limit laws can finish the calculation. If a factor has an indeterminate form such as infinity minus infinity, direct substitution is not a method; it is a signal to simplify further.

Working principle: Before taking a limit of a mixed expression, look for a useful factorization, conjugate, or standard limit form. Apply limit laws only after the expression has been put into a form where the factors have limits and any denominator is known to have a nonzero limit.

Standard Limits Used in Mixed Problems

The limits below are especially useful when an input tends to zero. They are stated here so that the transformations in the examples are explicit.

$$ \lim_{x\to 0}\frac{\ln(1+x)}{x}=1,\qquad \lim_{x\to 0}\frac{e^x-1}{x}=1,\qquad \lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0}\cos x=1. $$

For example, if \(c\) is a fixed nonzero real number, then \(c/n\to 0\), and substitution into the logarithm limit gives

$$ n\ln\left(1+\frac{c}{n}\right) = c\frac{\ln(1+c/n)}{c/n} \longrightarrow c. $$

The same kind of rewriting works for exponentials and sines:

$$ \frac{e^{c/n}-1}{c/n}\longrightarrow 1, \qquad \frac{\sin(c/n)}{c/n}\longrightarrow 1. $$

These are not new limit laws for arbitrary expressions. They are applications of the stated standard limits to inputs that approach zero. In a product, if each factor converges, the product converges to the product of the limits. In a quotient, if the numerator and denominator converge and the denominator limit is nonzero, the quotient converges to the quotient of those limits.

Worked Example: A Radical, a Logarithm, and a Sine Factor

Worked Example: A Radical, a Logarithm, and a Sine Factor

Evaluate

$$ a_n= \left(\sqrt{n^2+4n}-n\right) \left[n\ln\left(1+\frac{2}{n}\right)\right] \frac{\sin(3/n)}{3/n}, \qquad n\in\mathbb N. $$

The radical difference has the form infinity minus infinity, so simplify it before taking its limit. Multiply by the conjugate and divide by the same quantity:

$$ \sqrt{n^2+4n}-n = \frac{(\sqrt{n^2+4n}-n)(\sqrt{n^2+4n}+n)} {\sqrt{n^2+4n}+n} = \frac{4n}{\sqrt{n^2+4n}+n}. $$

Since \(n>0\), divide the numerator and denominator by \(n\):

$$ \frac{4n}{\sqrt{n^2+4n}+n} = \frac{4}{\sqrt{1+4/n}+1} \longrightarrow \frac{4}{1+1}=2. $$

For the logarithmic factor, make the argument \(2/n\) explicit:

$$ n\ln\left(1+\frac{2}{n}\right) = 2\frac{\ln(1+2/n)}{2/n} \longrightarrow 2. $$

The sine factor has limit \(1\), because \(3/n\to 0\) and \(\sin x/x\to 1\). Thus all three factors converge, and the product limit is

$$ \lim_{n\to\infty}a_n=2\cdot 2\cdot 1=4. $$

As a check, the simplified expression has factors tending to \(2\), \(2\), and \(1\), respectively; multiplying those values reproduces the claimed limit.

Why Conjugation Preserves the Limit

The key step in the first example is an exact identity, not an approximation. For any positive \(n\), the difference of square roots can be multiplied by its conjugate because the denominator \(\sqrt{n^2+4n}+n\) is positive and therefore not zero:

$$ (\sqrt{n^2+4n}-n)(\sqrt{n^2+4n}+n) =(n^2+4n)-n^2=4n. $$

Consequently, the original radical difference equals \(4n/(\sqrt{n^2+4n}+n)\) for every \(n\), not merely for large \(n\). Dividing top and bottom by \(n\) then exposes a denominator tending to \(2\), so the quotient limit law applies. This explains both why conjugation works and why it is useful: it changes a hard-to-read subtraction into a quotient with a nonzero limiting denominator.

Worked Example: Rational, Exponential, and Trigonometric Factors

Worked Example: Rational, Exponential, and Trigonometric Factors

Evaluate

$$ b_n= \frac{n^2}{n^2+3n+2} \frac{e^{2/n}-1}{2/n} \cos\left(\frac{1}{n}\right). $$

First divide the numerator and denominator of the rational factor by \(n^2\):

$$ \frac{n^2}{n^2+3n+2} = \frac{1}{1+3/n+2/n^2} \longrightarrow 1. $$

For the exponential factor, \(2/n\to 0\), so the standard exponential limit gives

$$ \frac{e^{2/n}-1}{2/n}\longrightarrow 1. $$

Also, \(1/n\to 0\), and continuity of cosine at zero gives \(\cos(1/n)\to\cos(0)=1\). Each factor has a finite limit, so

$$ \lim_{n\to\infty}b_n=1\cdot 1\cdot 1=1. $$

The check is direct: the rational factor approaches \(1\), the exponential ratio approaches \(1\), and the cosine factor approaches \(1\); their product therefore approaches \(1\).

Worked Example: A Logarithm Divided by a Radical Difference

Worked Example: A Logarithm Divided by a Radical Difference

Evaluate

$$ c_n= \frac{\sqrt n\,\ln(1+1/n)} {\sqrt{n+1}-\sqrt n}. $$

The denominator tends to zero, so the quotient law cannot be used by taking the limits of the numerator and denominator separately. Rationalize the denominator. Its conjugate is \(\sqrt{n+1}+\sqrt n\), and

$$ (\sqrt{n+1}-\sqrt n)(\sqrt{n+1}+\sqrt n) =(n+1)-n=1. $$

Therefore the expression is exactly

$$ c_n = \sqrt n\,\ln(1+1/n)(\sqrt{n+1}+\sqrt n) = n\ln(1+1/n) \left(\sqrt{1+1/n}+1\right). $$

The first factor tends to \(1\), by the logarithm limit with input \(1/n\):

$$ n\ln(1+1/n)=\frac{\ln(1+1/n)}{1/n}\longrightarrow 1. $$

The second factor tends to \(\sqrt{1}+1=2\), by continuity of the square-root function at \(1\). The product limit law now applies, giving

$$ \lim_{n\to\infty}c_n=1\cdot 2=2. $$

Substituting these factor limits into the transformed expression confirms the result: \(1\) times \(2\) is \(2\).

A Reliable Order of Work

Mixed problems become more manageable when the steps are chosen deliberately. The following procedure is a guide, not a rule that every problem must use every time.

1
Identify the pieces.
Separate rational, radical, logarithmic, exponential, and trigonometric factors when doing so makes their limits easier to see.
2
Simplify algebraic indeterminate forms.
For a difference of radicals, try a conjugate. For rational expressions, divide by the largest relevant power of \(n\).
3
Match special factors to standard limits.
Rewrite logarithmic, exponential, or sine expressions so the small input appears both inside and outside the function.
4
Check limit-law conditions.
Use a quotient law only after confirming that the denominator limit is nonzero. Use product laws once the individual factor limits exist.

Common Mistake: Substituting Infinity into a Difference

Consider the radical factor from the first example. A tempting but incorrect calculation is

$$ \sqrt{n^2+4n}-n \overset{\text{wrong}}{\longrightarrow} \infty-\infty=0. $$

This is wrong because \(\infty-\infty\) is not a defined arithmetic operation and does not determine a limit. In this example, the two terms grow at similar rates, and their difference approaches \(2\), not \(0\). The correction is to rationalize:

$$ \sqrt{n^2+4n}-n = \frac{4}{\sqrt{1+4/n}+1} \longrightarrow 2. $$

When direct substitution produces competing infinities or a zero denominator, do not assign a value to the indeterminate form. Rewrite the sequence until the limit laws apply to well-defined factors.

Why the Method Matters

Mixed limits show how several elementary ideas cooperate. Conjugation reveals the size of a radical difference; dividing by a dominant power simplifies rational expressions; and standard small-input limits handle logarithms, exponentials, and sines. The important skill is recognizing which transformation addresses the obstacle in each part.

Keep track of the conditions as well as the computations. A product law requires convergent factors, while a quotient law requires a denominator with a nonzero limit. An exact algebraic identity can be used before taking a limit, but an informal cancellation of terms that both grow without bound cannot.

Check Your Understanding

For each question, identify the useful simplification or standard limit before calculating.

  1. Evaluate \(\left(\sqrt{n^2+6n}-n\right)\frac{\sin(2/n)}{2/n}\).
  2. Find the limit of \(n\ln(1+5/n)\).
  3. Evaluate \(\frac{n^2+1}{2n^2-3n}\frac{e^{1/n}-1}{1/n}\).
  4. Why is it not valid to use the quotient limit law directly on \(\frac{\sqrt{n+1}-\sqrt n}{1/n}\) by treating the denominator as having limit zero?
  5. In a quotient of convergent sequences, what condition must the denominator limit satisfy before the quotient limit law applies?

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