Question

Use the table for each problem to find the given limits.

Answer 1

Answer:

1) $\lim_{x \to3 } (2f(x))+g(-x))=13$

2) $\lim_{x \to3 }\frac{g(x)}{f(-x)}=1/2$

Step-by-step explanation:

So we are given the limits:

$\lim_{x \to3 }f(x)=4\text{ and } \lim_{x \to-3 } f(x)=2$

And:

$\lim_{x \to 3 } g(x)= 1\text{ and } \lim_{x \to -3 } g(x)=5$

Question A)

We have the limit:

$\lim_{x \to3 } (2f(x))+g(-x))$

We can split this limit using our properties:

$= \lim_{x \to 3} (2f(x))+\lim_{x \to 3} g(-x)$

Now, use direct substitution. Substitute 3 for x. So:

$=2(f(3))+g(-3)$

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:

$=2(4)+(5)$

Multiply:

$=8+5=13$

So, our limit is:

$\lim_{x \to3 } (2f(x))+g(-x))=13$

Question B:

We have the limit:

$\lim_{x \to3 }\frac{g(x)}{f(-x)}$

Again, we can rewrite this as:

$\frac{\lim_{x \to3 }g(x)}{\lim_{x \to3 }f(-x)}}$

Direct substitution:

$=\frac{g(3)}{f(-3)}$

The value in the numerator, as given, is 1.

The value in the denominator will be 2. So:

$=1/2$

Therefore, our limit is:

$\lim_{x \to3 }\frac{g(x)}{f(-x)}=1/2$

And we're done!