Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

How to crack JEE Mains 2025 in 1 month? Follow study plan, revise high-score topics, experts advice
anam-khan

IIT JEE aspirant from Bihar found hanging in Kota hostel; 17th suicide this year
anam-khan

JEE Main 2025 just a month away; section-wise preparation tips, important topics, strategies

Latest Question