Question

if lim x->1 [f(x)+g(x)]=6 and lim x->1 f(x)xg(x)=4 then determine lim x->1 [f(x)-g(x)]^2. Please help and thank you!

Answer 1

there was a hint for this question and it says solve it as a system...

Answer 2

[f(x)-g(x)]^2 expand this and you can solve it using the laws of limit.

Answer 3

oh maybe they want you to solve the system

p+q = 6
p*q = 4

where p is the function value of f(x) when x = 1
q is the function value of g(x) when x = 1

Answer 4

\[\lim_{x \rightarrow 1} (f(x)+g(x)) = \lim_{x \rightarrow 1} f(x) + \lim_{x \rightarrow 1} g(x) \]

This kind of limit rule can help you with the problem

Answer 5

I guess we are suppose to just assume lim x->1 f(x) and/or lim x->1 g(x) exists?

Answer 6

That would be the only way to solve this question. You just need to break up the 3rd equation and calculate it using the 1st and the 2nd.

Answer 7

But we aren't given that.

Answer 8

Using jim_thompson5910's example,

If

p+q = 6
pq = 4

we can find (p-q)^2

Answer 9

how do you isolate the variables?

Answer 10

\[(f-g)^2=(f+g)^2-4(fg)\]

Answer 11

@ganeshie8 actually beat me in typing that

Answer 12

how did you get that?

Answer 13

\[\text{ we are given } f+g \\ \text{ so we can find } (f+g)^2 \\ \text{ but } (f-g)^2 \text{ is what we are looking for } (f-g)^2=f^2+g^2-2gf \\ \text{ but } (f+g)^2=f^2+g^2+2fg \text{ so subtracting } 4gf \text{ from that will give us } \\ (f+g)^2-4fg=(f-g)^2\]

Answer 14

\[\lim_{x \rightarrow 1}(f(x)+g(x))=6 \\ \text{ squaring both sides } [\lim_{x \rightarrow 1} f(x)+g(x)]^2=6^2 \\ \text{ using a limit property we can write this as } \lim_{x \rightarrow 1}(f(x)+g(x))^2=6^2 \]

Answer 15

\[\lim_{x \rightarrow 1}(f(x)-g(x))^2=\lim_{x \rightarrow 1}([f(x)]^2+[g(x)]^2-2 f(x)g(x)) \\ \text{ but we can't use what we are given for that } f^2+g^2 \text{ part } \\ \text{ so we have \to rewrite \in terms of what we are given } \\ =\lim_{x \rightarrow 1}([f(x)]^2+[g(x)]^2+2f(x)g(x)-2f(x)g(x)-2f(x)g(x)) \\ \text{ I added \in a zero here so I could do this: } \\ =\lim_{x \rightarrow 1}((f(x)+g(x))^2-2f(x)g(x)-2f(x)g(x)) \\ \text{ combine like terms there } \\ =\lim_{x \rightarrow 1}((f(x)+g(x))^2-4f(x)g(x))\]

Answer 16

\[=[\lim_{x \rightarrow 1}(f(x)+g(x))]^2-4 \cdot \lim_{x \rightarrow 1}f(x)g(x)\]
@esam2 look here
it is easier for me to type it here and make things look prettier :p

Answer 17

you are given the limit inside that square thing

Answer 18

you are given the limit of that one product also

Answer 19

if you dont mind can we start over?

Answer 20

\[(a-b)^2=a^2-2ab+b^2 \\ (a+b)^2=a^2+2ab+b^2 \]
are you familiar with these algebraic identities?

Answer 21

yes

Answer 22

\[\text{ \to express} (a-b)^2 \text{ \in terms of }(a+b)^2 \text{ we need \to add } 2ab \text{ but if we add } 2ab \\ \text{ we need \to also subtract } 2ab \\ \]
\[(a-b)^2=a^2-2ab+b^2+2ab-2ab \\ (a-b)^2=a^2+2ab+b^2-2ab-2ab \text{ move some terms around } \\ (a-b)^2=(a+b)^2-2ab-2ab \\ (a-b)^2=(a+b)^2-4ab \text{ combined like terms }\]

Answer 23

If you don't like this I think I can show you one more way how to get that equation
just subtract that one squared equation I gave you from the other
like so:
\[\text{ } \text{ } \text{ } \text{ } (a-b)^2=a^2-2ab+b^2 \\ -[(a+b)^2=a^2+2ab+b^2]\]
------------------------------
\[(a-b)^2-(a+b)^2=0-4ab+0 \\ (a-b)^2-(a+b)^2=-4ab \\ \text{ add } (a+b)^2 \text{ on both sides } (a-b)^2=-4ab+(a+b)^2 \]

Answer 24

oh i see. okay i think i got that part

Answer 25

\[=[\lim_{x \rightarrow 1}(f(x)+g(x))]^2-4 \cdot \lim_{x \rightarrow 1}f(x)g(x) \\ =(6)^2-4(4)\]

Answer 26

so the answer will be 20?

Answer 27

yep

Answer 28

thank you so much!!!

Answer 29

quick question what happen to the f(x) * g(x) =4?

Answer 30

we used it

Answer 31

\[=[\lim_{x \rightarrow 1}(f(x)+g(x))]^2-4 \cdot \lim_{x \rightarrow 1}f(x)g(x) \\ =(6)^2-\text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ }4 \cdot 4 \]
tried to line it up but i'm not the best at the latex code