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Find this limit with the help of the squeeze theorem.

If,x> 1, and x^2+5<g(x)<x^2+7, \lim _{x \rightarrow x} \frac{g(x)}{x^2-x+1}=?

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 hf(x) \leq g(x) \leq h(x) for \forall \mathrm{x}  in the domain of definition states the following: 

some real value of a, if \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>1 \quad \ldots(i) \\ & x^2+5<g(x)<x^2+7 \end{aligned} -------(ii)

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

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

Again the following is deduced.

x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0

Now from the inequality (ii) it follows as below

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

Apply the squeeze theorem to the inequality (iii).

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

\begin{aligned} & \frac{(1+0)}{(1-0+0)} \leq \lim _{x \rightarrow \infty} \frac{g(x)}{x^2-x+1} \leq \frac{(1+0)}{(1-0+0)} \\ & 1 \leq \lim _{x \rightarrow \infty} \frac{g(x)}{x^2-x+1} \leq 1 \end{aligned}

Therefore, the required value of the limit is 

\lim _{x \rightarrow \pi} \frac{g(x)}{x^2-x+1}=1

 

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mansi

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