# Limit Find Value Question

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 $\lim_{x\to a} f(x)=f(a)$ – Raymond Manzoni Jun 17 '12 at 22:27

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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How did you get h(x) = 4 that i dont understand? – soniccool Jun 17 '12 at 22:32
basically i understand how we got 2x5 but how did you get the 4 to subtract from it to get 6? – soniccool Jun 17 '12 at 22:33
@mystycs - It's in the problem statement. I edited my answer and added an intermediate step. – Ayman Hourieh Jun 17 '12 at 22:34
Houreih so because the limit they gave us =4 i used that 4 to subtract and get the 6? – soniccool Jun 17 '12 at 22:35
@mystycs - Happy to help! – Ayman Hourieh Jun 17 '12 at 22:41

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 ...

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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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Oh so becase 3 is 5 as f(x) i just put in 5 instead of 3 because i have 2 *3 but how did you get -4? – soniccool Jun 17 '12 at 22:29
Basically how did you get - 4 ? – soniccool Jun 17 '12 at 22:31
By hypothesis : $\lim(2f(x)-g(x))=4$. – JBC Jun 17 '12 at 22:35
So we used the 4 to subtract from 2x5 basically? – soniccool Jun 17 '12 at 22:36
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 $\lim(f+g)=l+l'$. – JBC Jun 17 '12 at 22:43