Question:
Let f(x) be a continuous function defined for all real numbers. Evaluate the following limit:
lim x → ∞ ( x 2 + 2 x − 3 x + 4 5 x + 1 ) \lim_{{x \to \infty}} \left(\sqrt{x^2 + 2x} - \frac{3x + 4}{5x + 1}\right) x → ∞ lim ( x 2 + 2 x − 5 x + 1 3 x + 4 ) Show all your work and justify each step using appropriate limit properties or theorems.
Answer:
To evaluate the given limit, we can simplify the expression and then apply limit properties step by step.
We start by applying algebraic simplification:
lim x → ∞ ( x 2 + 2 x − 3 x + 4 5 x + 1 ) \lim_{{x \to \infty}} \left(\sqrt{x^2 + 2x} - \frac{3x + 4}{5x + 1}\right) x → ∞ lim ( x 2 + 2 x − 5 x + 1 3 x + 4 ) = lim x → ∞ ( x 2 + 2 x ( 5 x + 1 ) − ( 3 x + 4 ) 5 x + 1 ) = \lim_{{x \to \infty}} \left(\frac{\sqrt{x^2 + 2x}(5x + 1) - (3x + 4)}{5x + 1}\right) = x → ∞ lim ( 5 x + 1 x 2 + 2 x ( 5 x + 1 ) − ( 3 x + 4 ) ) Next, we simplify the numerator by expanding:
= lim x → ∞ ( 5 x 3 + x x 2 + 2 x + x 2 + 2 x − 3 x − 4 5 x + 1 ) = \lim_{{x \to \infty}} \left(\frac{5x^3 + x\sqrt{x^2 + 2x} + \sqrt{x^2 + 2x} - 3x - 4}{5x + 1}\right) = x → ∞ lim ( 5 x + 1 5 x 3 + x x 2 + 2 x + x 2 + 2 x − 3 x − 4 ) Now we divide each term in the numerator by the highest power of x:
= lim x → ∞ ( 5 x 3 / x 3 + x 2 x 2 + 2 x / x 3 + x 2 + 2 x / x 3 − 3 x / x 3 − 4 / x 3 5 x / x 3 + 1 / x 3 ) = \lim_{{x \to \infty}} \left(\frac{5x^3/x^3 + x^2\sqrt{x^2 + 2x}/x^3 + \sqrt{x^2 + 2x}/x^3 - 3x/x^3 - 4/x^3}{5x/x^3 + 1/x^3}\right) = x → ∞ lim ( 5 x / x 3 + 1/ x 3 5 x 3 / x 3 + x 2 x 2 + 2 x / x 3 + x 2 + 2 x / x 3 − 3 x / x 3 − 4/ x 3 ) = lim x → ∞ ( 5 + x 2 + 2 x x + 1 x x 2 + 2 x − 3 x 2 − 4 x 3 5 x 2 + 1 x 3 ) = \lim_{{x \to \infty}} \left(\frac{5 + \frac{\sqrt{x^2 + 2x}}{x} + \frac{1}{x\sqrt{x^2 + 2x}} - \frac{3}{x^2} - \frac{4}{x^3}}{\frac{5}{x^2} + \frac{1}{x^3}}\right) = x → ∞ lim x 2 5 + x 3 1 5 + x x 2 + 2 x + x x 2 + 2 x 1 − x 2 3 − x 3 4 As x approaches infinity, every term with an x in the denominator approaches 0. Therefore, we can simplify further:
= lim x → ∞ ( 5 + x 2 + 2 x x + 1 x x 2 + 2 x − 3 x 2 − 4 x 3 ) lim x → ∞ ( 5 x 2 + 1 x 3 ) = \frac{\lim_{{x \to \infty}} (5 + \frac{\sqrt{x^2 + 2x}}{x} + \frac{1}{x\sqrt{x^2 + 2x}} - \frac{3}{x^2} - \frac{4}{x^3})}{\lim_{{x \to \infty}} (\frac{5}{x^2} + \frac{1}{x^3})} = lim x → ∞ ( x 2 5 + x 3 1 ) lim x → ∞ ( 5 + x x 2 + 2 x + x x 2 + 2 x 1 − x 2 3 − x 3 4 ) = 5 + 0 + 0 − 0 − 0 0 + 0 = \frac{5 + 0 + 0 - 0 - 0}{0 + 0} = 0 + 0 5 + 0 + 0 − 0 − 0 It appears that the denominator approaches 0 while the numerator approaches a non-zero value. Hence, this limit is of the form "infinity over zero", which can be an indeterminate form. To resolve this, we can apply L'Hôpital's Rule by taking the derivative of both the numerator and denominator with respect to x:
d d x ( 5 + x 2 + 2 x x + 1 x x 2 + 2 x − 3 x 2 − 4 x 3 ) / d d x ( 5 x 2 + 1 x 3 ) \frac{d}{dx}\left(5 + \frac{\sqrt{x^2 + 2x}}{x} + \frac{1}{x\sqrt{x^2 + 2x}} - \frac{3}{x^2} - \frac{4}{x^3}\right) / \frac{d}{dx}\left(\frac{5}{x^2} + \frac{1}{x^3}\right) d x d ( 5 + x x 2 + 2 x + x x 2 + 2 x 1 − x 2 3 − x 3 4 ) / d x d ( x 2 5 + x 3 1 ) Simplifying further, we get:
We can still apply L'Hôpital's Rule one more time to evaluate this limit. Taking the derivative again:
d 2 d x 2 ( 5 + x 2 + 2 x x + 1 x x 2 + 2 x − 3 x 2 − 4 x 3 ) / d 2 d x 2 ( 5 x 2 + 1 x 3 ) \frac{d^2}{dx^2}\left(5 + \frac{\sqrt{x^2 + 2x}}{x} + \frac{1}{x\sqrt{x^2 + 2x}} - \frac{3}{x^2} - \frac{4}{x^3}\right) / \frac{d^2}{dx^2}\left(\frac{5}{x^2} + \frac{1}{x^3}\right) d x 2 d 2 ( 5 + x x 2 + 2 x + x x 2 + 2 x 1 − x 2 3 − x 3 4 ) / d x 2 d 2 ( x 2 5 + x 3 1 ) After differentiating, we obtain:
( 2 ( x 2 + 2 x ) ) x 3 x 2 + 2 x / − 6 x 4 \frac{{\left(2(x^2+2x)\right)}}{x^3\sqrt{x^2+2x}} / \frac{{-6}}{{x^4}} x 3 x 2 + 2 x ( 2 ( x 2 + 2 x ) ) / x 4 − 6 Simplifying further yields:
− 12 ( x 2 + 2 x ) x 3 x 2 + 2 x \frac{{-12(x^2+2x)}}{{x^3\sqrt{x^2+2x}}} x 3 x 2 + 2 x − 12 ( x 2 + 2 x ) Now, we can evaluate the limit as x approaches infinity using this new expression:
lim x → ∞ − 12 ( x 2 + 2 x ) x 3 x 2 + 2 x \lim_{{x \to \infty}} \frac{{-12(x^2+2x)}}{{x^3\sqrt{x^2+2x}}} x → ∞ lim x 3 x 2 + 2 x − 12 ( x 2 + 2 x ) Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, the limit approaches 0:
Hence, the original limit lim x → ∞ ( x 2 + 2 x − 3 x + 4 5 x + 1 ) \lim_{{x \to \infty}} \left(\sqrt{x^2 + 2x} - \frac{3x + 4}{5x + 1}\right) lim x → ∞ ( x 2 + 2 x − 5 x + 1 3 x + 4 ) 0 .