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Find this limit \lim _{x \rightarrow 1} f(x), where,\frac{x^4-3 x^2-2}{3 x+2}<f(x)<\frac{x^4-3 x^2+2}{3 x+2}

Option: 1

\frac{x}{2}


Option: 2

\frac{1}{2}


Option: 3

\sqrt{2}x


Option: 4

Cannot be determined


Answers (1)

best_answer

Note the following important points:

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} g(x)=l

  • The Floor Function indicates the greatest integer function [GIF] denoted mathematically as [p] for only real values. This function [p] rounds downs the real number having any fractional or decimal part (if any) to the nearest integral value less than the indicated number.

The provided inequality is 

\frac{x^4-3 x^2-2}{3 x+2}<f(x)<\frac{x^4-3 x^2+2}{3 x+2} ----------(i)

Apply the following calculations to the inequality (i).

\begin{aligned} & \lim _{x \rightarrow 1} \frac{x^4-3 x^2-2}{3 x+2} \leq \lim _{x \rightarrow 1} f(x) \leq \lim _{x \rightarrow 1} \frac{x^4-3 x^2+2}{3 x+2} \\ & \frac{1^4-3 \times 1^2-2}{3 \times 1+2} \leq \lim _{x \rightarrow 1} f(x) \leq \frac{1^4-3 \times 1^2-2}{3 \times 1+2} \\ & \frac{1-3-2}{5} \leq \lim _{x \rightarrow 1} f(x) \leq \frac{1-3+2}{5} \\ & -\frac{4}{5} \leq \lim _{x \rightarrow 1} f(x) \leq 0 \end{aligned}

Note that the squeeze theorem cannot be applied here as the comparable limits -\frac{4}{5},0

unequal to each other.

Therefore, the required value of the limit cannot be determined.

 

 

Posted by

Pankaj Sanodiya

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