Question

Find the limits for the given functions:

(a) limx→2 x⁴ −2x³+x−2 / x³−2x²+4x−8
(b) lim x→0 x / sin(8x)

Answer 1

To find the limits of the given functions: (a) Simplify the function and evaluate the limit by direct substitution. For (b), use L'Hopital's Rule to evaluate the limit.

Step-by-step explanation:

To find the limits of the given functions:

(a) For the function ƒ(x) = (x⁴ - 2x³ + x - 2) / (x³ - 2x² + 4x - 8)

1. First, simplify the function by factoring out the common factors in the numerator and denominator: ƒ(x) = (x-2)(x³+x+1) / (x-2)(x²+4)
2. Next, cancel out the common factors of (x-2): ƒ(x) = x³+x+1 / x²+4
3. Now, you can evaluate the limit by directly substituting the value x=2 into the simplified function: limx→2ƒ(x) = 2³+2+1 / 2²+4 = 11/8

(b) For the function ƒ(x) = x / sin(8x)

1. As x approaches 0, the function sin(8x) approaches 0. Therefore, to find the limit, you can use the property that limx→a[ f(x) / g(x) ] = [ limx→a*f(x) ] / [ limx→ag(x) ], as long as the denominator is not zero.
2. Using this property, limx→0ƒ(x) = limx→0x / limx→0sin(8x) = 0 / 0
3. This is an indeterminate form, so you can apply L'Hopital's Rule. Taking the derivatives of both the numerator and denominator, you get limx→0ƒ(x) = limx→01 / 8cos(8x) = 1 / 8cos(0) = 1 / 8