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For all x≥0 if 4x≤g(x)≤2x^4−2x^2+4 for all x, evaluate lim x→1 g(x) as as x→1?

The aim of this question is to find the value of the given Limit of the functionThe basic concept behind this article is the understanding of the Limit Function and the Squeeze Theorem.

The Squeeze Theorem for the Limit Function is used where the given function is enclosed between two other functions. It is used to check if the limit of the function is correct by comparing it to two other functions with known limits.

As per the Squeeze Theorem:

\[f(x)\le\ g(x)\le\ h(x)\]

For the limit $x\rightarrow\ k$:

The limit of the function $g(x)$ is correct if:

\[f(k)=h(k)\]

Expert Answer

Given that:

\[4x\le\ g(x)\le2x^4-2x^2+4\]

This means that:

\[f(x)=4x\]

\[h(x)=2x^4-2x^2+4\]

The given limit is:

\[\ Limit=\lim_{x\rightarrow 1}\]

As per the Squeeze Theorem:

\[f(x)\le\ g(x)\le\ h(x)\]

For $x\rightarrow1$:

The limit of the function $g(x)$ is correct if:

\[f(1)=h(1)\]

So, for the function $f(x)$ at the given limit $x\rightarrow1$:

\[\lim_{x\rightarrow1}\ f(x)=4x\]

And:

\[f(1)=4(1)\]

\[f(1)=4\]

So, for the function $h(x)$ at the given limit $x\rightarrow1$:

\[\lim_{x\rightarrow1}\ h(x)=2x^4-2x^2+4\]

And:

\[h(1)=2{(1)}^4-2{(1)}^2+4\]

\[h(1)=2-2+4\]

\[h(1)=4\]

Hence, as per the above calculation, it is proved that:

\[\lim_{x\rightarrow1}\ f(x)=\lim_{x\rightarrow1}\ h(x)\]

Or:

\[f(1)=h(1)=4\]

So as per the Squeeze Theorem, if $f(1)=h(1)$, then the given limit is also correct for $g(x)$. Hence:

\[\lim_{x\rightarrow1}\ g(x)=\lim_{x\rightarrow1}\ f(x)=\ \lim_{x\rightarrow1}\ h(x)\]

And:

\[g(1)=f(1)=h(1)\]

\[g(1)=4=4\]

\[\lim_{x\rightarrow1}\ g(x)=g(1)=4\]

Numerical Result

For the given function $g(x)$ at the given limit $x\rightarrow1$, the value of $g(x)$ is:

\[\lim_{x\rightarrow1}\ g(x)=g(1)=4\]

Example

For $x\geq0$, find the value of limit $g(x)$ for the following squeezed function:

\[2x\ \le\ g\ (x)\ \le\ 2x^3\ +\ 2x\ -\ 2\]

Solution

Given that:

\[2x\ \le\ g\ (x)\ \le\ 2x^3\ +\ 2x\ -\ 2\]

This means that:

\[f\ (x)\ =\ 2x\]

\[h\ (x)\ =\ 2x^3\ +\ 2x\ -\ 2\]

The given limit is:

\[\ Limit\ =\ \lim_{x\rightarrow1}\]

As per the Squeeze Theorem:

\[f\ (x)\ \le\ g\ (x)\ \le\ h\ (x)\]

For $x\ \rightarrow\ 1$:

The limit of the function $g(x)$ is correct if:

\[f\ (1)\ =\ h\ (1)\]

So, for the function $f\ (x)$ at the given limit $x\ \rightarrow\ 1$:

\[\lim_{x\rightarrow1}\ \ f\ (x)\ =\ 2x\]

And:

\[f\ (1)\ =\ 2\ (1)\]

\[f\ (1)\ =\ 2\]

So, for the function $h\ (x)$ at the given limit $x\ \rightarrow\ 1$:

\[\lim_{x\rightarrow1}\ \ h\ (x)=\ \ 2x^3\ +\ 2x\ -\ 2\]

And:

\[h\ (1)=2{\ (1)}^3\ +\ 2\ (1)\ -\ 2\]

\[h\ (1)\ =\ 2\ +\ 2\ -\ 2\]

\[h\ (1)\ =\ 2\]

Hence as per the above calculation, it is proved that:

\[\lim_{x\rightarrow1}\ \ f\ (x)\ =\ \lim_{x\rightarrow1}\ \ h\ (\ x)\]

Or:

\[f\ (1)=h\ (1)=2\]

So as per the Squeeze Theorem, if $f(1)=h(1)$, then the given limit is also correct for $g(x)$. Hence:

\[\lim_{x\rightarrow1}\ g\ (x)\ =\ \lim_{x\rightarrow1}\ f\ (x)\ =\ \lim_{x\rightarrow1}\ h\ (x)\]

And:

\[g\ (1)\ =\ f\ (1)\ =\ h\ (1)\]

\[g\ (1)=\ 2\ =\ 2\]

\[\lim_{x\rightarrow1}\ g\ (x)\ =\ g\ (1)\ =\ 2\]

Hence, for the given function $g(x)$ at the given limit $x\ \rightarrow\ 1$, the value of $g(x)$ is:

\[\lim_{x\rightarrow1}\ g\ (x)\ =\ g\ (1)\ =\ 2\]

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