Problem to solve
Compute \[ \lim_{ x \to \infty } \frac{4x^5-1}{3x^3+7} \]
Solution
STEP 1
Find the following limit: lim_(x->∞) (4 x^5 - 1)/(3 x^3 + 7)
STEP 2
Hint: | Divide numerator and denominator of (4 x^5 - 1)/(3 x^3 + 7) by the denominator's leading term. The leading term in the denominator of (4 x^5 - 1)/(3 x^3 + 7) is x^3. Divide the numerator and denominator by this: lim_(x->∞) (4 x^2 - 1/x^3)/(3 + 7/x^3)
STEP 3
Hint: | Evaluate limits that tend to zero. The expressions -1/x^3 and 7/x^3 both tend to zero as x approaches ∞: lim_(x->∞) (4 x^2)/3
STEP 4
Hint: | Simplify the expression inside the limit. (4 x^2)/3 = (4 x^2)/3: lim_(x->∞) (4 x^2)/3
STEP 5
Hint: | If neither of the terms of a product approach 0 as x->a, then by the product rule, the limit of the product is the product of the limits. Applying the product rule, write lim_(x->∞) (4 x^2)/3 as (4 lim_(x->∞) x^2)/3: (4 lim_(x->∞) x^2)/3
STEP 6
Hint: | Apply the power rule. Using the power rule, write lim_(x->∞) x^2 as (lim_(x->∞) x)^2: (4 (lim_(x->∞) x)^2)/3
STEP 7
Hint: | As x approaches ∞, x approaches... lim_(x->∞) x = ∞: (4 ∞^2)/3
STEP 8
Hint: | Evaluate (4 ∞^2)/3. (4 ∞^2)/3 = ∞: Answer: | | ∞
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Edited by tododemate® - 2021