Question

Find the limit (calculus)

Answer 1

Answer:

I believe $\lim_{x \to a} f(x)$ also equals 0.

Step-by-step explanation:

We know that g(x) has to be greater than or equal to 0. That means f(x) has to be smaller than g(x) or equal to 0.

If we plug in any value of x, f(x) will be positive because of the absolute value. The only way we are going to get f(x) smaller than g(x) or equal to g(x) is to have very small fractions.

Therefore, if $\lim_{x \to a} g(x)=0$, then $\lim_{x \to a} f(x)=0$ because f(x) must be smaller or equal to 0 as well.

Answer 2

Answer:

$\displaystyle{ \lim_{x\to a}f(x)=0$

Step-by-step explanation:

The other user posted a great answer, but I would like to show you a more mathematical way!

We have $|f(x)|\leq g(x)$.

By the definition of absolute value, this is the same as saying:

$f(x)\leq g(x)\text{ and } -f(x)\leq g(x)$

For the right, we can divide both sides by -1 to acquire:

$f(x)\leq g(x)\text{ and } f(x)\geq- g(x)$

We can now combine them. This yields:

$-g(x)\leq f(x)\leq g(x)$

Now, we can use the Squeeze Theorem (a.k.a. Pinch or Sandwich Theorem). The Squeeze Theorem posits that if we have:

$g(x)\leq f(x)\leq h(x)$

And:

$\displaystyle{ \lim_{x \to c} g(x)= \lim_{x \to c} h(x)=L}$

Then the following must be true:

$\displaystyle{ \lim_{x \to c} f(x)=L$

We have:

$-g(x)\leq f(x)\leq g(x)$

And we know that:

$\displaystyle{ \lim_{x \to a} g(x)=0$

Then it follows that:

$\displaystyle{ \lim_{x \to a} -g(x)=- \lim_{x \to a} g(x)=0$

Therefore, by using the Squeeze Theorem, we can conclude that:

$\displaystyle{ \lim_{x\to a}f(x)=0$

And we're done!