Math / Calculus

Question5) $\lim _{x \rightarrow-\infty}-\frac{16 x}{x^{2}+16}$

Studdy Solution

STEP 1

What is this asking? We need to find the value that $- \frac{16x}{x^2 + 16}$ approaches as $x$ gets increasingly large in the negative direction (towards negative infinity). Watch out! Don't just plug in "negative infinity" directly.
We need to analyze the behavior of the expression as $x$ becomes very large and negative.

STEP 2

1. Rewrite the expression
2. Analyze the limit

STEP 3

To figure out what happens when $x$ gets super large, let's divide both the numerator and the denominator by the highest power of $x$ in the expression, which is $x^2$ .
This is a slick trick that helps us see what's really going on when $x$ is enormous!
Remember, dividing both the top and bottom by the same thing doesn't change the value of the fraction!

STEP 4

$-\frac{16x}{x^2 + 16} = -\frac{\frac{16x}{x^2}}{\frac{x^2}{x^2} + \frac{16}{x^2}} = -\frac{\frac{16}{x}}{1 + \frac{16}{x^2}}$ So, we've got a much nicer looking expression to work with now!

STEP 5

Now, let's see what happens to each part of our new expression as $x$ approaches negative infinity.
As $x$ gets super large (and negative), the fraction $\frac{16}{x}$ gets incredibly close to zero.
Similarly, $\frac{16}{x^2}$ also approaches zero as $x$ goes to negative infinity.
Why? Because $x^2$ becomes a gigantic positive number, making the whole fraction tiny!

STEP 6

Let's put it all together.
As $x$ approaches negative infinity: $-\frac{\frac{16}{x}}{1 + \frac{16}{x^2}} \rightarrow -\frac{0}{1 + 0} = -\frac{0}{1} = 0$

STEP 7

The limit of the expression as $x$ approaches negative infinity is 0.

Was this helpful?

Question5) $\lim _{x \rightarrow-\infty}-\frac{16 x}{x^{2}+16}$

Studdy Solution

STEP 1

STEP 2

1. Rewrite the expression
2. Analyze the limit

STEP 3

STEP 4

$-\frac{16x}{x^2 + 16} = -\frac{\frac{16x}{x^2}}{\frac{x^2}{x^2} + \frac{16}{x^2}} = -\frac{\frac{16}{x}}{1 + \frac{16}{x^2}}$ So, we've got a much nicer looking expression to work with now!

STEP 5

STEP 6

Let's put it all together.
As $x$ approaches negative infinity: $-\frac{\frac{16}{x}}{1 + \frac{16}{x^2}} \rightarrow -\frac{0}{1 + 0} = -\frac{0}{1} = 0$

STEP 7

The limit of the expression as $x$ approaches negative infinity is 0.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question5) lim⁡x→−∞−16xx2+16\lim _{x \rightarrow-\infty}-\frac{16 x}{x^{2}+16}limx→−∞​−x2+1616x​

Studdy Solution

STEP 1

STEP 2

1. Rewrite the expression2. Analyze the limit

STEP 3

STEP 4

−16xx2+16=−16xx2x2x2+16x2=−16x1+16x2-\frac{16x}{x^2 + 16} = -\frac{\frac{16x}{x^2}}{\frac{x^2}{x^2} + \frac{16}{x^2}} = -\frac{\frac{16}{x}}{1 + \frac{16}{x^2}}−x2+1616x​=−x2x2​+x216​x216x​​=−1+x216​x16​​ So, we've got a much nicer looking expression to work with now!

STEP 5

STEP 6

Let's put it all together.As xxx approaches negative infinity: −16x1+16x2→−01+0=−01=0-\frac{\frac{16}{x}}{1 + \frac{16}{x^2}} \rightarrow -\frac{0}{1 + 0} = -\frac{0}{1} = 0−1+x216​x16​​→−1+00​=−10​=0

STEP 7

The limit of the expression as xxx approaches negative infinity is **0**.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question5) $\lim _{x \rightarrow-\infty}-\frac{16 x}{x^{2}+16}$

1. Rewrite the expression
2. Analyze the limit

$-\frac{16x}{x^2 + 16} = -\frac{\frac{16x}{x^2}}{\frac{x^2}{x^2} + \frac{16}{x^2}} = -\frac{\frac{16}{x}}{1 + \frac{16}{x^2}}$ So, we've got a much nicer looking expression to work with now!

Let's put it all together.
As $x$ approaches negative infinity: $-\frac{\frac{16}{x}}{1 + \frac{16}{x^2}} \rightarrow -\frac{0}{1 + 0} = -\frac{0}{1} = 0$

The limit of the expression as $x$ approaches negative infinity is 0.