Verifying limits of sequences

Exercises · Limits of sequences · with worked solutions · PDF

Exercise 1

Using the definition of limit, verify that:

\[ \displaystyle \lim_{n\to+\infty}\frac{n}{2n+5}=\frac{1}{2} \]

Solution

We have

\[ a_{n}=\frac{n}{2n+5} \]

We must verify that for every \(\varepsilon>0\) there exists \(n(\varepsilon) \in \N\) such that

\[ n>n(\varepsilon) \Rightarrow \frac{1}{2}-\varepsilon<a_{n}<\frac{1}{2}+\varepsilon \]

We have

\[ a_{n}-\frac{1}{2}=-\frac{5}{4n+10}, \]

hence we must verify that

\[ n>n(\varepsilon)\Rightarrow -\varepsilon<-\frac{5}{4n+10}<\varepsilon \]

The second inequality is always true, while the first is equivalent to

\[ 4n+10>\frac{5}{\varepsilon} {\rm ~~~hence~it~is~satisfied~for~~} n>\frac{5}{4\varepsilon}-\frac{5}{2} \]

For a fixed \(\varepsilon > 0\), it suffices to choose the first integer

\[ n(\varepsilon) > \frac{5}{4\varepsilon}-\frac{5}{2} \]

to satisfy the condition required by the definition of limit.

Exercise 2

Using the definition of limit, verify that:

\[ \displaystyle \lim_{n\to+\infty}\frac{1}{\sqrt{n+1}}=0 \]

Solution

We have

\[ a_{n}=\frac{1}{\sqrt{n+1}} \]

We must verify that for every \(\varepsilon>0\) there exists \(n(\varepsilon) \in \N\) such that

\[ n>n(\varepsilon) \Rightarrow -\varepsilon<a_{n}< \varepsilon \]

The first inequality is always true, while the second is equivalent to

\[ \sqrt{n+1}>\frac{1}{\varepsilon} {\rm ~~~hence~it~is~satisfied~for~~} n>\frac{1}{\varepsilon^{2}}-1 \]

For a fixed \(\varepsilon > 0\), it suffices to choose the first integer

\[ n(\varepsilon) > \frac{1}{\varepsilon^{2}}-1 \]

to satisfy the condition required by the definition of limit.

Exercise 3

Using the definition of limit, verify that:

\[ \displaystyle \lim_{n\to+\infty}\log{\left(1+\frac{1}{n}\right)}=0 \]

Solution

We have

\[ a_{n}=\log{\left(1+\frac{1}{n}\right)} \]

We must verify that for every \(\varepsilon>0\) there exists \(n(\varepsilon) \in \N\) such that

\[ n>n(\varepsilon) \Rightarrow -\varepsilon<a_{n}< \varepsilon \]

The first inequality is always true, while the second is equivalent to

\[ 1+\frac{1}{n}<e^{\varepsilon} {\rm ~~~hence~it~is~satisfied~for~~} n>\frac{1}{e^{\varepsilon}-1} \]

For a fixed \(\varepsilon > 0\), it suffices to choose the first integer

\[ n(\varepsilon) > \frac{1}{e^{\varepsilon}-1} \]

to satisfy the condition required by the definition of limit.

Exercise 4

Using the definition of limit, verify that:

\[ \displaystyle \lim_{n\to+\infty}\sqrt{4+\frac{1}{n}}=2 \]

Solution

We have

\[ a_{n}=\sqrt{4+\frac{1}{n}} \]

We must verify that for every \(\varepsilon>0\) there exists \(n(\varepsilon) \in \N\) such that

\[ n>n(\varepsilon) \Rightarrow 2-\varepsilon<a_{n}<2+\varepsilon \]

The first inequality is always true since \(a_n > 2\) for every \(n\), while the second is equivalent to

\[ 4+\frac{1}{n}<4+\varepsilon^{2}+4\varepsilon {\rm ~~~hence~it~is~satisfied~for~~} n>\frac{1}{\varepsilon^{2}+4\varepsilon} \]

For a fixed \(\varepsilon > 0\), it suffices to choose the first integer

\[ n(\varepsilon) > \frac{1}{\varepsilon^{2}+4\varepsilon} \]

to satisfy the condition required by the definition of limit.

Exercise 5

Using the definition of limit, verify that:

\[ \displaystyle \lim_{n\to+\infty}(n^{2}-1)=+\infty \]

Solution

We have

\[ a_{n}=n^{2}-1 \]

We must verify that for every \(M>0\) there exists \(n(M) \in \N\) such that

\[ n> n(M) \Rightarrow a_{n}>M \]

The inequality

\[ n^{2}-1>M {\rm ~~~is~satisfied~for~~} n>\sqrt{M+1} \]

Hence, for a fixed \(M > 0\), it suffices to choose the first integer

\[ n(M) > \sqrt{M+1} \]

to satisfy the required divergence condition.

Exercise 6

Using the definition of limit, verify that:

\[ \displaystyle \lim_{n\to+\infty}\log{(\sqrt{n}+1)}=+\infty \]

Solution

We have

\[ a_{n}=\log{(\sqrt{n}+1)} \]

We must verify that for every \(M>0\) there exists \(n(M) \in \N\) such that

\[ n>n(M)\Rightarrow a_{n}>M \]

The inequality

\[ \log{(\sqrt{n}+1)}>M {\rm ~~~is~satisfied~for~~} n>(e^{M}-1)^2 \]

Hence, for a fixed \(M > 0\), it suffices to choose the first integer

\[ n(M) > (e^{M}-1)^2 \]

to satisfy the required divergence condition.