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If $f$ and $g$ are continuous functions, with $f(3) = 5$ and

$\lim_{x \to 3} (2f(x) − g(x)) = 4$

find $g(3)$.

I am confused at how to tackle this question, I understand I have to find $g(3)$ but do I plus in $3$ for $x$? How do I go about getting the solution because apparently someone got $g(3) = 6$ as the solution?

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    hint: for continuous functions $\lim_{x\to a} f(x)=f(a)$2012-06-17

3 Answers 3

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Define $h(x)$ as: $ h(x) = 2f(x) - g(x) $

We can write $g(x)$ in terms of $f(x)$ and $h(x)$ as: $ g(x) = 2f(x) - h(x) $

Since $f(x)$ is continuous: $ \lim_{x \to 3} f(x) = f(3) = 5 $

From the problem statement, we have: $ \lim_{x \to 3} h(x) = \lim_{x \to 3} \left(2f(x) - g(x)\right)= 4 $

Therefore: $ \lim_{x \to 3} g(x) = \lim_{x \to 3} \left(2f(x) - h(x)\right) = 2 \times 5 - 4 = 6 $

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    @mystycs - Happy to help!2012-06-17
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We have $g(x)=2f(x)-(2f(x)-g(x))$, hence, by continuity, $g(3)=\lim_{x\rightarrow 3}g(x)=2\times \lim f(x)-\lim(2f(x)-g(x))=2\times5-4=6$

Remark : if $f$ is continuous, $g(a)=\lim_{x\rightarrow a}g(x)$

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    We used $g(3)=\lim_{x\rightarrow 3}g(x)$ by continuity and elementary operations on limits : if $\lim f=l\in\mathbb R$ then $\lim\lambda f=\lambda l$ and if $\lim f=l$ and $\lim g=l'$ then $\lim(f+g)=l+l'$.2012-06-17
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We have: $\lim_{x \to 3} (2f(x) − g(x)) =4$

When $f$ and $g$ are continuous functions:

$\lim_{x \to 3} (2f(x) − g(x)) =2f(3)-g(3)$ $\Downarrow$ $2f(3)-g(3)=4$

$f(3) = 5$, so $g(3)$ is ...