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40. If \( 2 x \leqslant g(x) \leqslant x^{4}-x^{2}+2 \) for all \( x \), evaluate \( \lim _{x \rightarrow 1} g(x) \).

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Answer:

To evaluate the limit of \(g(x)\) as \(x\) approaches 1, we can use the squeeze theorem (also known as the sandwich theorem). The squeeze theorem states that if you have three functions, \(f(x)\), \(g(x)\), and \(h(x)\), and for all \(x\) in some interval except possibly at a point \(c\), the inequality \(f(x) \leq g(x) \leq h(x)\) holds, and the limits of \(f(x)\) and \(h(x)\) as \(x\) approaches \(c\) are both equal to \(L\), then the limit of \(g(x)\) as \(x\) approaches \(c\) is also \(L\).

In this case, we have:

\[2x \leq g(x) \leq x^4 - x^2 + 2\]

Now, let's evaluate the limits of the bounding functions as \(x\) approaches 1:

1. \(\lim_{x \to 1} (2x) = 2\cdot 1 = 2\)

2. \(\lim_{x \to 1} (x^4 - x^2 + 2) = 1^4 - 1^2 + 2 = 1 - 1 + 2 = 2\)

So, both \(2x\) and \(x^4 - x^2 + 2\) have a limit of 2 as \(x\) approaches 1.

Now, by the squeeze theorem, since \(2x \leq g(x) \leq x^4 - x^2 + 2\) and both \(2x\) and \(x^4 - x^2 + 2\) have a limit of 2 as \(x\) approaches 1, the limit of \(g(x)\) as \(x\) approaches 1 is also 2.

Therefore,

\[\lim_{x \to 1} g(x) = 2\]

Explanation:

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