Studdy AI Logo
Math  /  Calculus

Questionlim⁡x→05x3+8x23x4−16x2\lim _{x \rightarrow 0} \frac{5 x^{3}+8 x^{2}}{3 x^{4}-16 x^{2}}

Studdy Solution

STEP 1

What is this asking? We need to find the limit of a fraction of polynomials as xx approaches 0. Watch out! We can't just plug in x=0x = 0 directly, because we'd get 0 divided by 0, which is undefined!

STEP 2

1. Simplify the expression
2. Substitute the value

STEP 3

We can **factor out** x2x^2 from both the numerator and the denominator.
This is a great idea because it helps us simplify the expression and get rid of the pesky division by zero issue! 5x3+8x23x4−16x2=x2(5x+8)x2(3x2−16) \frac{5x^3 + 8x^2}{3x^4 - 16x^2} = \frac{x^2(5x + 8)}{x^2(3x^2 - 16)}

STEP 4

Since xx is approaching 0 (but *not equal* to 0), x2x^2 is not zero.
So, we can divide both the numerator and the denominator by x2x^2.
This simplifies our expression even further!
Remember, anything divided by itself is 1! x2(5x+8)x2(3x2−16)=5x+83x2−16 \frac{x^2(5x + 8)}{x^2(3x^2 - 16)} = \frac{5x + 8}{3x^2 - 16}

STEP 5

Now that we've simplified our expression, we can **substitute** x=0x = 0 into the new expression.
This is safe to do now because we won't get a division by zero error! lim⁡x→05x+83x2−16=5⋅0+83⋅02−16 \lim_{x \rightarrow 0} \frac{5x + 8}{3x^2 - 16} = \frac{5 \cdot 0 + 8}{3 \cdot 0^2 - 16}

STEP 6

Let's simplify this expression to get our **final result**: 5⋅0+83⋅02−16=8−16=−12 \frac{5 \cdot 0 + 8}{3 \cdot 0^2 - 16} = \frac{8}{-16} = -\frac{1}{2}

STEP 7

The limit of the expression as xx approaches 0 is −12-\frac{1}{2}.

Was this helpful?

Studdy solves anything!

banner

Start learning now

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

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord