In this collection we present 20 practice problems on operations with limits of sequences. The exercises illustrate how to correctly apply the rules for sums, differences, products, and quotients, paying close attention to the assumptions of the theorems and to indeterminate forms.
The guiding idea is the following: if two sequences converge to finite limits, then, under suitable assumptions, the limit of the sequence obtained by an algebraic operation is computed by applying the same operation to the limits.
Exercise 1 — level ★☆☆☆☆
Find the limit of the sequence
\[ c_n=\frac{1}{n}+\frac{n}{n+1}. \]
Answer
\[ \lim_{n\to+\infty}\left(\frac{1}{n}+\frac{n}{n+1}\right)=1. \]
Solution
The sequence is the sum of two sequences:
\[ \frac{1}{n} \qquad\text{and}\qquad \frac{n}{n+1}. \]
We compute the two limits separately.
We have
\[ \lim_{n\to+\infty}\frac{1}{n}=0. \]
Moreover,
\[ \frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}. \]
Since
\[ \frac{1}{n+1}\to0, \]
it follows that
\[ \frac{n}{n+1}\to1. \]
We may therefore apply the rule for the limit of a sum:
\[ \lim_{n\to+\infty}\left(\frac{1}{n}+\frac{n}{n+1}\right) = \lim_{n\to+\infty}\frac{1}{n} + \lim_{n\to+\infty}\frac{n}{n+1}. \]
Hence
\[ \lim_{n\to+\infty}\left(\frac{1}{n}+\frac{n}{n+1}\right)=0+1=1. \]
Exercise 2 — level ★☆☆☆☆
Find the limit of the sequence
\[ c_n=\left(2+\frac{1}{n}\right)-\left(3-\frac{2}{n}\right). \]
Answer
\[ \lim_{n\to+\infty}\left[\left(2+\frac{1}{n}\right)-\left(3-\frac{2}{n}\right)\right]=-1. \]
Solution
Consider the two sequences
\[ a_n=2+\frac{1}{n} \qquad\text{and}\qquad b_n=3-\frac{2}{n}. \]
Since
\[ \frac{1}{n}\to0, \]
we obtain
\[ a_n=2+\frac{1}{n}\to2. \]
Likewise, since
\[ \frac{2}{n}\to0, \]
we have
\[ b_n=3-\frac{2}{n}\to3. \]
We now apply the rule for the limit of a difference:
\[ \lim_{n\to+\infty}(a_n-b_n) = \lim_{n\to+\infty}a_n-\lim_{n\to+\infty}b_n. \]
Thus
\[ \lim_{n\to+\infty}\left[\left(2+\frac{1}{n}\right)-\left(3-\frac{2}{n}\right)\right]=2-3=-1. \]
Exercise 3 — level ★☆☆☆☆
Find the limit of the sequence
\[ c_n=\left(2+\frac{1}{n}\right)\left(3-\frac{1}{n}\right). \]
Answer
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n}\right)\left(3-\frac{1}{n}\right)=6. \]
Solution
The sequence is the product of two sequences:
\[ a_n=2+\frac{1}{n}, \qquad b_n=3-\frac{1}{n}. \]
Since
\[ \frac{1}{n}\to0, \]
we have
\[ a_n=2+\frac{1}{n}\to2. \]
Moreover,
\[ b_n=3-\frac{1}{n}\to3. \]
We may apply the rule for the limit of a product, since both sequences have finite limits:
\[ \lim_{n\to+\infty}(a_n b_n) = \left(\lim_{n\to+\infty}a_n\right) \left(\lim_{n\to+\infty}b_n\right). \]
Therefore
\[ \lim_{n\to+\infty}\left(2+\frac{1}{n}\right)\left(3-\frac{1}{n}\right)=2\cdot3=6. \]
Exercise 4 — level ★☆☆☆☆
Find the limit of the sequence
\[ c_n=\frac{2+\displaystyle \frac{1}{n}}{5-\displaystyle \frac{3}{n}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{2+\displaystyle \frac{1}{n}}{5-\displaystyle \frac{3}{n}}=\frac25. \]
Solution
We examine the numerator and the denominator separately.
For the numerator:
\[ 2+\frac{1}{n}\to2. \]
For the denominator:
\[ 5-\frac{3}{n}\to5. \]
The limit of the denominator is \(5\), hence it is nonzero. We may therefore apply the rule for the limit of a quotient.
We obtain
\[ \lim_{n\to+\infty}\frac{2+\displaystyle \frac{1}{n}}{5-\displaystyle \frac{3}{n}} = \frac{\lim_{n\to+\infty}\left(2+\displaystyle \frac{1}{n}\right)} {\lim_{n\to+\infty}\left(5-\displaystyle \frac{3}{n}\right)} = \frac25. \]
The condition on the denominator is essential: here it is satisfied, because the limit of the denominator is \(5\neq0\).
Exercise 5 — level ★★☆☆☆
Find the limit of the sequence
\[ c_n=\frac{3n+1}{n}. \]
Answer
\[ \lim_{n\to+\infty}\frac{3n+1}{n}=3. \]
Solution
We split the fraction:
\[ \frac{3n+1}{n} = \frac{3n}{n}+\frac{1}{n}. \]
Thus
\[ \frac{3n+1}{n}=3+\frac{1}{n}. \]
Now we use the operations on limits. The constant sequence \(3\) tends to \(3\), while
\[ \frac{1}{n}\to0. \]
By the rule for the limit of a sum we obtain
\[ \lim_{n\to+\infty}\left(3+\frac{1}{n}\right) = 3+0=3. \]
Therefore
\[ \lim_{n\to+\infty}\frac{3n+1}{n}=3. \]
Exercise 6 — level ★★☆☆☆
Find the limit of the sequence
\[ c_n=\frac{4n-5}{2n+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{4n-5}{2n+1}=2. \]
Solution
The numerator and denominator are first-degree polynomials in \(n\). We divide both by \(n\):
\[ \frac{4n-5}{2n+1} = \frac{4-\displaystyle \frac{5}{n}}{2+\displaystyle \frac{1}{n}}. \]
We now compute the limit of the numerator:
\[ 4-\frac{5}{n}\to4. \]
We compute the limit of the denominator:
\[ 2+\frac{1}{n}\to2. \]
The limit of the denominator is \(2\), hence nonzero. We may apply the quotient rule for limits:
\[ \lim_{n\to+\infty}\frac{4-\displaystyle \frac{5}{n}}{2+\displaystyle \frac{1}{n}} = \frac{4}{2}=2. \]
Hence
\[ \lim_{n\to+\infty}\frac{4n-5}{2n+1}=2. \]
Exercise 7 — level ★★☆☆☆
Find the limit of the sequence
\[ c_n=\left(\frac{n}{n+1}\right)\left(\frac{2n+3}{n}\right). \]
Answer
\[ \lim_{n\to+\infty}\left(\frac{n}{n+1}\right)\left(\frac{2n+3}{n}\right)=2. \]
Solution
The sequence is the product of two sequences:
\[ a_n=\frac{n}{n+1} \qquad\text{and}\qquad b_n=\frac{2n+3}{n}. \]
For the first sequence:
\[ \frac{n}{n+1}=1-\frac{1}{n+1}\to1. \]
For the second:
\[ \frac{2n+3}{n}=2+\frac{3}{n}\to2. \]
Both have finite limits. We may therefore apply the rule for the limit of a product:
\[ \lim_{n\to+\infty}\left(\frac{n}{n+1}\right)\left(\frac{2n+3}{n}\right) = 1\cdot2=2. \]
Exercise 8 — level ★★☆☆☆
Find the limit of the sequence
\[ c_n=\left(\frac{n+1}{n}\right)^2. \]
Answer
\[ \lim_{n\to+\infty}\left(\frac{n+1}{n}\right)^2=1. \]
Solution
We rewrite the base of the power:
\[ \frac{n+1}{n}=1+\frac1n. \]
Since
\[ \frac1n\to0, \]
we obtain
\[ 1+\frac1n\to1. \]
The given sequence is
\[ \left(1+\frac1n\right)^2 = \left(1+\frac1n\right)\left(1+\frac1n\right). \]
We apply the rule for the limit of a product:
\[ \lim_{n\to+\infty}\left(1+\frac1n\right)^2 = 1\cdot1=1. \]
Exercise 9 — level ★★☆☆☆
Find the limit of the sequence
\[ c_n=\frac{\displaystyle \frac{2}{n}+3}{4-\displaystyle \frac{1}{n}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{2}{n}+3}{4-\displaystyle \frac{1}{n}}=\frac34. \]
Solution
We examine the numerator and the denominator.
The numerator is
\[ \frac{2}{n}+3. \]
Since
\[ \frac{2}{n}\to0, \]
we have
\[ \frac{2}{n}+3\to3. \]
The denominator is
\[ 4-\frac{1}{n}. \]
Since
\[ \frac{1}{n}\to0, \]
we have
\[ 4-\frac{1}{n}\to4. \]
The limit of the denominator is nonzero. We may therefore apply the rule for the limit of a quotient:
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{2}{n}+3}{4-\displaystyle \frac{1}{n}}=\frac{3}{4}. \]
Exercise 10 — level ★★★☆☆
Find the limit of the sequence
\[ c_n=\frac{n^2+3n}{2n^2-5}. \]
Answer
\[ \lim_{n\to+\infty}\frac{n^2+3n}{2n^2-5}=\frac12. \]
Solution
The numerator and denominator are polynomials of the same degree, namely degree \(2\).
We divide the numerator and denominator by \(n^2\):
\[ \frac{n^2+3n}{2n^2-5} = \frac{1+\displaystyle \frac3n}{2-\displaystyle \frac5{n^2}}. \]
We compute the limits of the individual parts:
\[ \frac3n\to0 \qquad\text{and}\qquad \frac5{n^2}\to0. \]
Thus the numerator tends to
\[ 1+0=1, \]
while the denominator tends to
\[ 2-0=2. \]
Since the limit of the denominator is \(2\neq0\), we may apply the rule for the limit of a quotient:
\[ \lim_{n\to+\infty}\frac{1+\displaystyle \frac3n}{2-\displaystyle \frac5{n^2}}=\frac12. \]
Therefore
\[ \lim_{n\to+\infty}\frac{n^2+3n}{2n^2-5}=\frac12. \]
Exercise 11 — level ★★★☆☆
Find the limit of the sequence
\[ c_n=\frac{2n^2-n+1}{n^2+4}. \]
Answer
\[ \lim_{n\to+\infty}\frac{2n^2-n+1}{n^2+4}=2. \]
Solution
We divide the numerator and denominator by \(n^2\), that is, by the highest power of \(n\) that appears:
\[ \frac{2n^2-n+1}{n^2+4} = \frac{2-\displaystyle \frac1n+\displaystyle \frac1{n^2}}{1+\displaystyle \frac4{n^2}}. \]
Now observe that
\[ \frac1n\to0 \qquad\text{and}\qquad \frac1{n^2}\to0. \]
Thus the numerator tends to
\[ 2-0+0=2, \]
and the denominator tends to
\[ 1+0=1. \]
Since the limit of the denominator is \(1\neq0\), we apply the rule for the limit of a quotient:
\[ \lim_{n\to+\infty}\frac{2-\displaystyle \frac1n+\displaystyle \frac1{n^2}}{1+\displaystyle \frac4{n^2}} = \frac21=2. \]
Hence
\[ \lim_{n\to+\infty}\frac{2n^2-n+1}{n^2+4}=2. \]
Exercise 12 — level ★★★☆☆
Find the limit of the sequence
\[ c_n=\frac{3n+1}{n^2+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{3n+1}{n^2+1}=0. \]
Solution
The denominator has higher degree than the numerator. We divide the numerator and denominator by \(n^2\):
\[ \frac{3n+1}{n^2+1} = \frac{\displaystyle \frac3n+\displaystyle \frac1{n^2}}{1+\displaystyle \frac1{n^2}}. \]
Now
\[ \frac3n\to0, \qquad \frac1{n^2}\to0. \]
Thus the numerator tends to
\[ 0+0=0, \]
while the denominator tends to
\[ 1+0=1. \]
Since the limit of the denominator is nonzero, we may apply the quotient rule:
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac3n+\displaystyle \frac1{n^2}}{1+\displaystyle \frac1{n^2}} = \frac01=0. \]
Hence
\[ \lim_{n\to+\infty}\frac{3n+1}{n^2+1}=0. \]
Exercise 13 — level ★★★☆☆
Find the limit of the sequence
\[ c_n=\frac{n^3+2n}{n^2+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{n^3+2n}{n^2+1}=+\infty. \]
Solution
The numerator has degree \(3\), whereas the denominator has degree \(2\). We therefore expect the quotient to grow without bound.
We divide the numerator and denominator by \(n^2\):
\[ \frac{n^3+2n}{n^2+1} = \frac{n+\displaystyle \frac2n}{1+\displaystyle \frac1{n^2}}. \]
As \(n\to+\infty\), the numerator
\[ n+\frac2n \]
tends to \(+\infty\), because the term \(n\) grows without bound.
The denominator, on the other hand, tends to
\[ 1+0=1. \]
Hence the quotient tends to \(+\infty\).
We can also give a lower estimate. For \(n\geq1\), we have
\[ n^2+1\leq2n^2 \]
and
\[ n^3+2n\geq n^3. \]
Therefore
\[ \frac{n^3+2n}{n^2+1}\geq \frac{n^3}{2n^2}=\frac n2. \]
Since
\[ \frac n2\to+\infty, \]
the given sequence also tends to \(+\infty\).
Exercise 14 — level ★★★☆☆
Find the limit of the sequence
\[ c_n=\frac{1}{n}\left(4-\frac{3}{n}\right). \]
Answer
\[ \lim_{n\to+\infty}\frac{1}{n}\left(4-\frac{3}{n}\right)=0. \]
Solution
The sequence is a product:
\[ \frac{1}{n} \qquad\text{and}\qquad 4-\frac{3}{n}. \]
The first factor tends to \(0\):
\[ \frac1n\to0. \]
The second factor tends to \(4\), because
\[ \frac3n\to0. \]
Hence
\[ 4-\frac3n\to4. \]
We may apply the rule for the limit of a product:
\[ \lim_{n\to+\infty}\frac{1}{n}\left(4-\frac{3}{n}\right)=0\cdot4=0. \]
In this case there is no indeterminate form: the second factor tends to a finite number, not to \(+\infty\).
Exercise 15 — level ★★★☆☆
Find the limit of the sequence
\[ c_n=\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}+1}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}+1}=0. \]
Solution
We examine the numerator and the denominator.
The numerator is
\[ \frac1n, \]
so it tends to \(0\).
The denominator is
\[ \frac1n+1, \]
and it tends to
\[ 0+1=1. \]
The limit of the denominator is \(1\), hence nonzero. We may therefore apply the rule for the limit of a quotient:
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}+1} = \frac{0}{1}=0. \]
Exercise 16 — level ★★★★☆
Find the limit of the sequence
\[ c_n=\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}}=1. \]
Solution
The numerator tends to \(0\):
\[ \frac1n\to0. \]
The denominator also tends to \(0\):
\[ \frac1n\to0. \]
We therefore cannot apply the theorem on the limit of a quotient directly, since the required hypothesis is that the limit of the denominator be nonzero.
The expression is of the type
\[ \frac{0}{0}, \]
that is, an indeterminate form.
Nevertheless, we can simplify the sequence. For every positive integer \(n\), we have
\[ \frac{\displaystyle \frac1n}{\displaystyle \frac1n}=1. \]
Hence the sequence is constantly equal to \(1\).
Therefore
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n}}=1. \]
This exercise shows that a form \(\displaystyle \frac00\) does not automatically determine the limit: one must examine the expression.
Exercise 17 — level ★★★★☆
Find the limit of the sequence
\[ c_n=\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n^2}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n^2}}=+\infty. \]
Solution
The numerator tends to \(0\):
\[ \frac1n\to0. \]
The denominator also tends to \(0\):
\[ \frac1{n^2}\to0. \]
We are therefore facing a form of the type
\[ \frac00. \]
We cannot apply the quotient rule directly, since the limit of the denominator is \(0\).
We simplify the expression:
\[ \frac{\displaystyle \frac1n}{\displaystyle \frac1{n^2}} = \frac1n\cdot n^2. \]
Hence
\[ \frac{\displaystyle \frac1n}{\displaystyle \frac1{n^2}}=n. \]
Since
\[ n\to+\infty, \]
we obtain
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n}}{\displaystyle \frac{1}{n^2}}=+\infty. \]
This confirms that the form \(\displaystyle \frac00\) is indeterminate: in a previous exercise it gave \(1\), whereas here it gives \(+\infty\).
Exercise 18 — level ★★★★☆
Find the limit of the sequence
\[ c_n=\frac{\displaystyle \frac{1}{n^2}}{\displaystyle \frac{1}{n}}. \]
Answer
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n^2}}{\displaystyle \frac{1}{n}}=0. \]
Solution
In this case too, the numerator and denominator both tend to \(0\):
\[ \frac1{n^2}\to0 \qquad\text{and}\qquad \frac1n\to0. \]
The expression is thus a form of the type
\[ \frac00. \]
We cannot apply the quotient rule directly. We must simplify.
We write:
\[ \frac{\displaystyle \frac1{n^2}}{\displaystyle \frac1n} = \frac1{n^2}\cdot n. \]
Hence
\[ \frac{\displaystyle \frac1{n^2}}{\displaystyle \frac1n} = \frac1n. \]
Since
\[ \frac1n\to0, \]
we conclude that
\[ \lim_{n\to+\infty}\frac{\displaystyle \frac{1}{n^2}}{\displaystyle \frac{1}{n}}=0. \]
This is a further example of the fact that the form \(\frac00\) can yield different results.
Exercise 19 — level ★★★★★
Find the limit of the sequence
\[ c_n=\sqrt{n^2+n}-n. \]
Answer
\[ \lim_{n\to+\infty}\left(\sqrt{n^2+n}-n\right)=\frac12. \]
Solution
The sequence is
\[ c_n=\sqrt{n^2+n}-n. \]
Observe that
\[ \sqrt{n^2+n}\to+\infty \qquad\text{and}\qquad n\to+\infty. \]
Thus the expression is of the type
\[ +\infty-\infty, \]
that is, an indeterminate form. We cannot compute the limit by simply subtracting the limits.
To remove the indeterminacy, we rationalize:
\[ \sqrt{n^2+n}-n = \frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}. \]
In the numerator we use the difference of squares:
\[ (\sqrt{n^2+n})^2-n^2=n^2+n-n^2=n. \]
Hence
\[ \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}. \]
Now we factor out \(n\) inside the radical:
\[ \sqrt{n^2+n}=\sqrt{n^2\left(1+\frac1n\right)}. \]
Since \(n>0\), we have
\[ \sqrt{n^2\left(1+\frac1n\right)} = n\sqrt{1+\frac1n}. \]
Thus
\[ \frac{n}{\sqrt{n^2+n}+n} = \frac{n}{n\sqrt{1+\frac1n}+n}. \]
Factoring out \(n\) in the denominator:
\[ \frac{n}{n\left(\sqrt{1+\frac1n}+1\right)} = \frac{1}{\sqrt{1+\frac1n}+1}. \]
Now
\[ \frac1n\to0, \]
so
\[ \sqrt{1+\frac1n}\to\sqrt1=1. \]
Therefore
\[ \frac{1}{\sqrt{1+\frac1n}+1}\to\frac{1}{1+1}=\frac12. \]
We conclude that
\[ \lim_{n\to+\infty}\left(\sqrt{n^2+n}-n\right)=\frac12. \]
Exercise 20 — level ★★★★★
Find the limit of the sequence
\[ c_n=n\left(\frac{n+1}{n}-1\right). \]
Answer
\[ \lim_{n\to+\infty}n\left(\frac{n+1}{n}-1\right)=1. \]
Solution
Consider the sequence
\[ c_n=n\left(\frac{n+1}{n}-1\right). \]
Looking at the factors separately, we have
\[ n\to+\infty \]
and
\[ \frac{n+1}{n}-1\to1-1=0. \]
The expression is thus of the type
\[ +\infty\cdot0, \]
that is, an indeterminate form. We cannot automatically conclude that the limit is \(0\) or \(+\infty\).
We must simplify the expression. We first compute the quantity in parentheses:
\[ \frac{n+1}{n}-1 = \frac{n+1}{n}-\frac{n}{n} = \frac{1}{n}. \]
Hence
\[ c_n=n\cdot\frac1n. \]
Therefore
\[ c_n=1 \]
for every positive integer \(n\).
The sequence is thus constantly equal to \(1\). Consequently,
\[ \lim_{n\to+\infty}n\left(\frac{n+1}{n}-1\right)=1. \]
This exercise shows that the form \(\infty\cdot0\) is also indeterminate: one must transform the expression before computing the limit.