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  • Find the limit  <img alt="\lim _{n \rightarrow \infty} \sqrt{n^2+n}-n" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%5Csqrt%7Bn%5E2&p

Find the limit  \lim _{n \rightarrow \infty} \sqrt{n^2+n}-n  with the help of the squeeze theorem, where  n\epsilon N 

Option: 1

\pm \frac{1}{2}


Option: 2

1


Option: 3

0


Option: 4

\frac{1}{2}


Answers (1)

best_answer

The Sandwich theorem also known as squeeze play theorem applicable for any three real valued functions f(x) g(x) and h(x) that shares the same domain such that  f(x)\leq g(x)\leq h(x) for \forall x in the domain of definition states the following: For some real value of a, \lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x) then \lim _{x \rightarrow a} f(x)=l 

The provided limit is

\lim _{n \rightarrow \infty} \sqrt{n^2+n}-n --------------------(i)

 Note the following calculation:

\begin{aligned} & \sqrt{n^2+n}-n \\ & =\frac{\left(\sqrt{n^2+n}-n\right)\left(\sqrt{n^2+n}+n\right)}{\left(\sqrt{n^2+n}+n\right)} \\ & =\frac{\left(\sqrt{\left(n^2+n\right)}\right)^2-(n)^2}{\left(\sqrt{n^2\left(1+\frac{1}{n}\right)}+n\right)} \\ & \end{aligned}

\begin{aligned} & =\frac{n^2+n-n^2}{n\left(\sqrt{\left(1+\frac{1}{n}\right)}+1\right)} \\ & \therefore \sqrt{n^2+n}-n=\frac{1}{\left(\sqrt{\left(1+\frac{1}{n}\right)}+1\right)} \end{aligned} --------------------(ii)

Now, consider the following observations:

  • n \in N, \sqrt{\frac{n+1}{n}} \geq 1

 

  • \sqrt{\frac{n+1}{n}}=\sqrt{1+\frac{1}{n}}=\left(1+\frac{1}{n}\right)^{\frac{1}{2}} \leq\left(1+\frac{1}{2 n}\right)
  • \therefore 1 \leq \sqrt{1+\frac{1}{n}} \leq\left(1+\frac{1}{2 n}\right)

Compare the equation (ii) and the inequality (iii).

\begin{aligned} & 1 \leq \sqrt{1+\frac{1}{n}} \leq\left(1+\frac{1}{2 n}\right) \\ & (1+1) \leq\left(\sqrt{1+\frac{1}{n}}+1\right) \leq\left(\left(1+\frac{1}{2 n}\right)+1\right) \\ & 2 \leq\left(\sqrt{1+\frac{1}{n}}-1\right) \leq 2+\frac{1}{2 n} \end{aligned}

\begin{aligned} & \frac{1}{2} \geq \frac{1}{\left.\sqrt{1+\frac{1}{n}}+1\right)^2 2+\frac{1}{2 n}} \\ & \frac{1}{2+\frac{1}{2 n}} \leq \frac{1}{\left(\sqrt{1}\left(1+\frac{1}{n}+1\right)\right.} \leq \frac{1}{2} \end{aligned}

Apply the squeeze theorem to the inequality (iv) and use the equation (ii).

\begin{aligned} & \lim _{n \rightarrow \infty}\left(\frac{1}{2+\frac{1}{2 n}}\right) \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+\frac{1}{n}}+1}\right) \leq \lim _{n \rightarrow \infty} \frac{1}{2} \\ & \frac{1}{2} \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+\frac{1}{n}}+1}\right) \leq \frac{1}{2} \end{aligned}

\frac{1}{2} \leq \lim _{n \rightarrow \infty}\left(\sqrt{n^2+n}-n\right) \leq \frac{1}{2}

Therefore, the required value of the limit is 

\lim _{n \rightarrow \infty} \sqrt{n^2+n}-n=\frac{1}{2}

 

 

 

Posted by

SANGALDEEP SINGH

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