Question

3. Find the following limits: (a) $\lim_{x\to\infty} \frac{6x - 8}{3x^2 - 2x - 4}$ (b) $\lim_{x\to\infty} \frac{6x^2 - 8}{3x^2 - 2x - 4}$ (c) $\lim_{x\to\infty} \frac{6x^3 - 8}{3x^2 - 2x - 4}$ (d) $\lim_{x\to\infty} \frac{4x - 8}{\sqrt{x^2 + 16}}$ a = b = c = d = (e) $\lim_{x\to 3} \frac{x^2 - 3x}{x^2 + x - 12}$ (f) $\lim_{x\to 2} \frac{x^2 - x - 2}{x - 2}$ e = f =

          3. Find the following limits:
(a) $\lim_{x\to\infty} \frac{6x - 8}{3x^2 - 2x - 4}$ (b) $\lim_{x\to\infty} \frac{6x^2 - 8}{3x^2 - 2x - 4}$ (c) $\lim_{x\to\infty} \frac{6x^3 - 8}{3x^2 - 2x - 4}$ (d) $\lim_{x\to\infty} \frac{4x - 8}{\sqrt{x^2 + 16}}$
a = ____ b = ____ c = ____ d = ____
(e) $\lim_{x\to 3} \frac{x^2 - 3x}{x^2 + x - 12}$ (f) $\lim_{x\to 2} \frac{x^2 - x - 2}{x - 2}$
e = ____ f = ____

3. Find the following limits:
(a) limx→∞(6x - 8)/(3x^2 - 2x - 4) (b) limx→∞(6x^2 - 8)/(3x^2 - 2x - 4) (c) limx→∞(6x^3 - 8)/(3x^2 - 2x - 4) (d) limx→∞(4x - 8)/(√(x^2 + 16))
a = b = c = d =
(e) limx→ 3(x^2 - 3x)/(x^2 + x - 12) (f) limx→ 2(x^2 - x - 2)/(x - 2)
e = f =

Added by Montserrat C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Carson Merrill

07/02/2021

Step 1

Therefore, the limit is: \lim_(x->\infty )(6x-8)/(3x^(2)-2x-4) = \lim_(x->\infty )(6x-8)/(3x^(2)) = \lim_(x->\infty )(6x/3x^(2)) = \lim_(x->\infty )(2/x) = 0 Show more…

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Find the following limits: (a) lim_(x->infty )(6x-8)/(3x^(2)-2x-4) (b) lim_(x->infty )(6x^(2)-8)/(3x^(2)-2x-4) (c) lim_(x->infty )(6x^(3)-8)/(3x^(2)-2x-4) (d) lim_(x->-infty )(4x-8)/(sqrt(x^(2)+16)) a=,b= c=,d= (e) lim_(x->3)(x^(2)-3x)/(x^(2)+x-12) (f) lim_(x->2)(x^(2)-x-2)/(x-2) e= f= 3. Find the following limits: 6x3-8 4x-8 6x-8 6x2-8 (c) lim (d) lim a lim b lim 3x2-2x-4 x-3x2-2x-4 8 Vx2+16 x-3x2-2x-4 d= b= C= a= x2-x-2 (f) lim x->2 x-2 x2-3x (e) lim x->3x+x-12 f= e=