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Find this limit with the help of the squeeze theorem. 

if,  x> 0, 2 x+5<f(x)<2 x+7 \quad \lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2}=?

Option: 1

4


Option: 2

0


Option: 3

-1


Option: 4

1


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(\mathrm{x}) \leq g(\mathrm{x}) \leq h(\mathrm{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} g(x)=l

The following are the stated conditions:

\begin{aligned} & x>0 \quad \ldots(i) \\ & 2 x+5<f(x)<2 x+7 \end{aligned}---------(ii)

From the inequality (i), the following is evident.

\begin{aligned} & 2 x+5>0 \\ & 2 x+7>0 \end{aligned}

Now from the inequality (ii) it follows as below.

\begin{aligned} & (2 x+5)^2<\{f(x)\}^2<(2 x+7)^2 \\ & \frac{(2 x+5)^2}{(x-1)^2}<\frac{\{f(x)\}^2}{(x-1)^2}<\frac{(2 x+7)^2}{(x-1)^2} \end{aligned} --------(iii)

Apply the squeeze theorem to the inequality (iii).

\begin{aligned} & \lim _{x \rightarrow \infty} \frac{(2 x+5)^2}{(x-1)^2} \leq \lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2} \leq \lim _{x \rightarrow \infty} \frac{(2 x+7)^2}{(x-1)^2} \\ & \lim _{x \rightarrow \infty} \frac{\left(2+\frac{5}{x}\right)^2}{\left(1-\frac{1}{x}\right)^2} \leq \lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2} \leq \lim _{x \rightarrow \infty} \frac{\left(2+\frac{7}{x}\right)^2}{\left(1-\frac{1}{x}\right)^2} \end{aligned}

\begin{aligned} & \frac{\left(\lim _{x \rightarrow \infty}\left(2+\frac{5}{x}\right)\right)^2}{\left(\lim _{x \rightarrow \infty}\left(1-\frac{1}{x}\right)\right)^2} \leq \lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2} \leq \frac{\left(\lim _{x \rightarrow \infty}\left(2+\frac{7}{x}\right)\right)^2}{\left(\lim _{x \rightarrow \infty}\left(1-\frac{1}{x}\right)\right)^2} \\ & \frac{(2+0)^2}{(1-0)^2} \leq \lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2} \leq \frac{(2+0)^2}{(1-0)^2} \\ & 4 \leq \lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2} \leq 4 \end{aligned}

Therefore, the required value of the limit is 

\lim _{x \rightarrow \infty} \frac{\{f(x)\}^2}{(x-1)^2}=4

Posted by

manish painkra

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