Problem to solve
Compute \[ \lim_{ x \to -\infty } \frac{4x-1}{\sqrt{x^2+2}} \]
Solution
STEP 1
Find the following limit: lim_(x->-∞) (4 x - 1) (x^2 + 2)^(-1/2)
STEP 2
Hint: | Factor a square root out of the quotient. For large negative values of x: lim_(x->-∞) (4 x - 1)/sqrt(x^2 + 2) = lim_(x->-∞)-sqrt((4 x - 1)^2/(x^2 + 2)) = -lim_(x->-∞) sqrt((4 x - 1)^2/(x^2 + 2)): -lim_(x->-∞)(sqrt((4 x - 1)^2/(x^2 + 2)))
STEP 3
Hint: | Using the power rule, pass the limit through the 2nd root. lim_(x->-∞) sqrt((4 x - 1)^2/(x^2 + 2)) = sqrt(lim_(x->-∞) (4 x - 1)^2/(x^2 + 2)): sqrt(lim_(x->-∞) (4 x - 1)^2/(x^2 + 2))
STEP 4
Hint: | Expand the numerator of (4 x - 1)^2/(x^2 + 2). (4 x - 1)^2 = 16 x^2 - 8 x + 1: -sqrt(lim_(x->-∞) (16 x^2 - 8 x + 1)/(x^2 + 2))
STEP 5
Hint: | Divide numerator and denominator of (16 x^2 - 8 x + 1)/(x^2 + 2) by the denominator's leading term. The leading term in the denominator of (16 x^2 - 8 x + 1)/(x^2 + 2) is x^2. Divide the numerator and denominator by this: -sqrt(lim_(x->-∞) (16 - 8/x + 1/x^2)/(1 + 2/x^2))
STEP 6
Hint: | Evaluate limits that tend to zero. The expressions 1/x^2, -8/x and 2/x^2 all tend to zero as x approaches -∞: -sqrt(16)
STEP 7
Hint: | Evaluate -sqrt(16). -sqrt(16) = -4: Answer: | | -4
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Edited by tododemate® - 2021