Answer:
1) [tex]\lim_{x \to3 } (2f(x))+g(-x))=13[/tex]
2) [tex]\lim_{x \to3 }\frac{g(x)}{f(-x)}=1/2[/tex]
Step-by-step explanation:
So we are given the limits:
[tex]\lim_{x \to3 }f(x)=4\text{ and } \lim_{x \to-3 } f(x)=2[/tex]
And:
[tex]\lim_{x \to 3 } g(x)= 1\text{ and } \lim_{x \to -3 } g(x)=5[/tex]
Question A)
We have the limit:
[tex]\lim_{x \to3 } (2f(x))+g(-x))[/tex]
We can split this limit using our properties:
[tex]= \lim_{x \to 3} (2f(x))+\lim_{x \to 3} g(-x)[/tex]
Now, use direct substitution. Substitute 3 for x. So:
[tex]=2(f(3))+g(-3)[/tex]
We are given that f(3) (or the limit as x approaches towards 3) is 4.
We know that the limit as x tends towards -3 of g(x) is 5. In other words, g(-3) can be said to be 5. So:
[tex]=2(4)+(5)[/tex]
Multiply:
[tex]=8+5=13[/tex]
So, our limit is:
[tex]\lim_{x \to3 } (2f(x))+g(-x))=13[/tex]
Question B:
We have the limit:
[tex]\lim_{x \to3 }\frac{g(x)}{f(-x)}[/tex]
Again, we can rewrite this as:
[tex]\frac{\lim_{x \to3 }g(x)}{\lim_{x \to3 }f(-x)}}[/tex]
Direct substitution:
[tex]=\frac{g(3)}{f(-3)}[/tex]
The value in the numerator, as given, is 1.
The value in the denominator will be 2. So:
[tex]=1/2[/tex]
Therefore, our limit is:
[tex]\lim_{x \to3 }\frac{g(x)}{f(-x)}=1/2[/tex]
And we're done!
