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  • Find the limit <img alt="\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+8 \frac{1}{\sqrt{n+n^2}}\right)" src="https://entrancecorner.onc

Find the limit \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+8 \frac{1}{\sqrt{n+n^2}}\right) with the help of the squeeze theorem,

where n\epsilon N.

Option: 1

\pm \pi


Option: 2

1


Option: 3

0


Option: 4

\pi


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,  if 

\lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x) then \lim _{x \rightarrow a} f(x)=l 

Note the following observations:

\begin{aligned} & 1<m<n ; \quad n, m \in N ; m \neq 0 \\ & 1+n^2<m+n^2<n+n^2 \\ & \sqrt{1+n^2}<\sqrt{m+n^2}<\sqrt{n+n^2} \\ & \frac{1}{\sqrt{1+n^2}}>\frac{1}{\sqrt{m+n^2}}>\frac{1}{\sqrt{n+n^2}} \\ & \frac{1}{\sqrt{n+n^2}}<\frac{1}{\sqrt{m+n^2}}<\frac{1}{\sqrt{1+n^2}} \end{aligned} -------------------------(i)

put m=1,2,3...n in the in the equality (i) to get the following.

\begin{aligned} & \frac{1}{\sqrt{n+n^2}}<\frac{1}{\sqrt{1+n^2}}<\frac{1}{\sqrt{1+n^2}} \\ & \frac{1}{\sqrt{n+n^2}}<\frac{1}{\sqrt{2+n^2}}<\frac{1}{\sqrt{1+n^2}} \\ & \text { ? } \\ & \frac{1}{\sqrt{n+n^2}}<\frac{1}{\sqrt{n+n^2}}<\frac{1}{\sqrt{1+n^2}} \end{aligned} 

\begin{aligned} & \frac{n}{\sqrt{n+n^2}}<\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+1 \frac{1}{\sqrt{n+n^2}}\right)<\frac{n}{\sqrt{1+n^2}} \\ & \frac{1}{\sqrt{1+\frac{1}{n}}}<\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+8 \frac{1}{\sqrt{n+n^2}}\right)<\frac{1}{\sqrt{1+\frac{1}{n^2}}} \end{aligned}

\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+\frac{1}{n}}}\right) \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+1 \frac{1}{\sqrt{n+n^2}}\right) \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+\frac{1}{n^2}}}\right)

\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+0}}\right) \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+\varnothing \frac{1}{\sqrt{n+n^2}}\right) \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+0}}\right)

1 \leq \lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+\frac{1}{\sqrt{n+n^2}}\right) \leq 1

Apply the squeeze theorem to the inequality (ii).

Therefore, the required value of the limit is

\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{1+n^2}}+\frac{1}{\sqrt{2+n^2}}+\frac{1}{\sqrt{3+n^2}}+8 \frac{1}{\sqrt{n+n^2}}\right)=1

Posted by

himanshu.meshram

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