Consider the following limit.

$\lim _{x \rightarrow-\infty} \frac{6 x}{\sqrt{\dot{x}^2+6}}$

(a) Use the definition of limits at infinity to find the value of $N$ that corresponds to $\varepsilon=0.5$.
$N=$

(b) Use the definition of limits at infinity to find the value of $N$ that corresponds to $\varepsilon=0.1$.
$N=$

Simplify the expression x 2 + 6 ​ 6 x ​ as x → − ∞ , obtaining 1 + x 2 6 ​ ​ − 6 ​ .
Use the definition of limits at infinity to set up the inequality ​ 6 − 1 + x 2 6 ​ ​ 6 ​ ​ < ϵ .
Solve the inequality for x to find N = − ( 6 − ϵ 6 ​ ) 2 − 1 6 ​ ​ .
Substitute ϵ = 0.5 and ϵ = 0.1 into the expression for N to get N ≈ − 5.618 ​ and N ≈ − 13.248 ​ , respectively.

Explanation

Problem Analysis We are given the limit lim x → − ∞ ​ x 2 + 6 ​ 6 x ​ and we need to find the value of N that corresponds to ε = 0.5 and ε = 0.1 using the definition of limits at infinity. This means we want to find N such that if x < N , then ​ x 2 + 6 ​ 6 x ​ − ( − 6 ) ​ < ϵ .

Simplifying the Expression First, let's simplify the expression inside the absolute value. Since x → − ∞ , we can assume x < 0 . Then x 2 ​ = ∣ x ∣ = − x . So, x 2 + 6 ​ = x 2 ( 1 + x 2 6 ​ ) ​ = ∣ x ∣ 1 + x 2 6 ​ ​ = − x 1 + x 2 6 ​ ​ . Thus, we have \begin{align*} \left|\frac{6 x}{\sqrt{x^2+6}} - (-6)\right| &= \left|\frac{6 x}{\sqrt{x^2+6}} + 6\right| \ &= \left|\frac{6 x}{-x\sqrt{1+\frac{6}{x^2}}} + 6\right| \ &= \left|\frac{-6}{\sqrt{1+\frac{6}{x^2}}} + 6\right| \ &= \left|6 - \frac{6}{\sqrt{1+\frac{6}{x^2}}}\right| \end{align*}

Solving the Inequality We want to find N such that if x < N , then ​ 6 − 1 + x 2 6 ​ ​ 6 ​ ​ < ϵ . This is equivalent to


− ϵ < 6 − 1 + x 2 6 ​ ​ 6 ​ < ϵ
Isolating the term with x , we get
6 − ϵ < 1 + x 2 6 ​ ​ 6 ​ < 6 + ϵ
Taking the reciprocal,
6 + ϵ 1 ​ < 6 1 + x 2 6 ​ ​ ​ < 6 − ϵ 1 ​
Multiplying by 6,
6 + ϵ 6 ​ < 1 + x 2 6 ​ ​ < 6 − ϵ 6 ​
Squaring the inequality,
( 6 + ϵ 6 ​ ) 2 < 1 + x 2 6 ​ < ( 6 − ϵ 6 ​ ) 2
Subtracting 1,
( 6 + ϵ 6 ​ ) 2 − 1 < x 2 6 ​ < ( 6 − ϵ 6 ​ ) 2 − 1
Taking the reciprocal,
( 6 − ϵ 6 ​ ) 2 − 1 1 ​ < 6 x 2 ​ < ( 6 + ϵ 6 ​ ) 2 − 1 1 ​
Multiplying by 6,
( 6 − ϵ 6 ​ ) 2 − 1 6 ​ < x 2 < ( 6 + ϵ 6 ​ ) 2 − 1 6 ​
Taking the square root,
( 6 − ϵ 6 ​ ) 2 − 1 6 ​ ​ < ∣ x ∣ < ( 6 + ϵ 6 ​ ) 2 − 1 6 ​ ​
Since x < 0 , we have
− ( 6 + ϵ 6 ​ ) 2 − 1 6 ​ ​ < x < − ( 6 − ϵ 6 ​ ) 2 − 1 6 ​ ​
Therefore, N = − ( 6 − ϵ 6 ​ ) 2 − 1 6 ​ ​ .

Calculating N for given epsilon (a) For ϵ = 0.5 , we have

N = − ( 6 − 0.5 6 ​ ) 2 − 1 6 ​ ​ = − ( 5.5 6 ​ ) 2 − 1 6 ​ ​ = − ( 11 12 ​ ) 2 − 1 6 ​ ​ = − 121 144 ​ − 1 6 ​ ​ = − 121 23 ​ 6 ​ ​ = − 23 6 ⋅ 121 ​ ​ = − 23 726 ​ ​ ≈ − 31.565 ​ ≈ − 5.618
(b) For ϵ = 0.1 , we have
N = − ( 6 − 0.1 6 ​ ) 2 − 1 6 ​ ​ = − ( 5.9 6 ​ ) 2 − 1 6 ​ ​ = − ( 59 60 ​ ) 2 − 1 6 ​ ​ = − 3481 3600 ​ − 1 6 ​ ​ = − 3481 119 ​ 6 ​ ​ = − 119 6 ⋅ 3481 ​ ​ = − 119 20886 ​ ​ ≈ − 175.5126 ​ ≈ − 13.248

Final Answer Therefore, for (a), N ≈ − 5.618 and for (b), N ≈ − 13.248 .

Examples
Understanding limits at infinity is crucial in fields like physics and engineering. For example, when analyzing the behavior of electrical circuits as time approaches infinity, engineers use limits to determine the steady-state current or voltage. Similarly, in astrophysics, limits at infinity help describe the gravitational potential of a star or galaxy at large distances. These concepts allow scientists and engineers to make accurate predictions and design systems that function reliably under extreme conditions.

Answered by GinnyAnswer | 2025-07-05