Value of $\displaystyle \lim_{x \to 0} g(x)$ Given Limit Statements

[karush]

$\textsf{find the value that $\displaystyle \lim_{x \to 0} g(x)$ must have if the
given limit statements hold.}$
$$\displaystyle \lim_{x \to 0} \left(\frac{4-g(x)}{x} \right)=1$$

OK the only answer I saw by observation was 2 but the book says it is 4
not sure how you get it with steps

 

[Chris L T521]

$\textsf{find the value that $\displaystyle \lim_{x \to 0} g(x)$ must have if the
given limit statements hold.}$
$$\displaystyle \lim_{x \to 0} \left(\frac{4-g(x)}{x} \right)=1$$

OK the only answer I saw by observation was 2 but the book says it is 4
not sure how you get it with steps

2 doesn't work; if $\displaystyle\lim_{x\to 0} g(x)=2$, then $\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x} \rightarrow \frac{2}{0}$, which is undefined.

For $\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x}=1$, $x$ must be a factor of $4-g(x)$. Suppose $4-g(x) = x f(x)$, where $g(x)$ and $f(x)$ both have limits as $x\to 0$. Then

\[\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x} = \lim_{x\to 0}f(x) = 1.\]

Furthermore, note that $\displaystyle\lim_{x\to 0}(4-g(x)) = \lim_{x\to 0}x f(x)$. Since $\displaystyle\lim_{x\to 0}f(x)=1$, we find that

\[\lim_{x\to 0}(4-g(x)) =\lim_{x\to 0}xf(x) \implies 4-\lim_{x\to 0}g(x) = 0\cdot 1 \implies \lim_{x\to 0}g(x) = 4.\]

I hope this makes sense!

 

[Greg]

Alternatively, recall your work with indeterminate forms. What value must $g(x)$ take to obtain $\frac00$ in the limit statement?

 

[karush]

2 doesn't work; if $\displaystyle\lim_{x\to 0} g(x)=2$, then $\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x} \rightarrow \frac{2}{0}$, which is undefined.
For $\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x}=1$, $x$ must be a factor of $4-g(x)$. Suppose $4-g(x) = x f(x)$, where $g(x)$ and $f(x)$ both have limits as $x\to 0$. Then
\[\displaystyle\lim_{x\to 0}\frac{4-g(x)}{x} = \lim_{x\to 0}f(x) = 1.\]
Furthermore, note that $\displaystyle\lim_{x\to 0}(4-g(x)) = \lim_{x\to 0}x f(x)$. Since $\displaystyle\lim_{x\to 0}f(x)=1$, we find that
\[\lim_{x\to 0}(4-g(x)) =\lim_{x\to 0}xf(x) \implies 4-\lim_{x\to 0}g(x) = 0\cdot 1 \implies \lim_{x\to 0}g(x) = 4.\]
I hope this makes sense!

yes thank you for the expanded explanation the book was to short on the subject