Math Problem Statement

Determine the following limits: a) lim (x -> -∞) x / (x+1)^2, b) lim (x -> -∞) x^2 / (x^2 + 1), c) lim (x -> -∞) e^x / x, d) lim (x -> ∞) arctan(-x)e^x, e) lim (x -> ∞) e^(x/2), f) lim (x -> ∞) e^(2x) / x.

Solution

Let's solve the given limits one by one.

a) limxx(x+1)2\lim_{x \to -\infty} \frac{x}{(x + 1)^2}

As xx \to -\infty:

  • The numerator is xx \to -\infty.
  • The denominator is (x+1)2x2(x + 1)^2 \approx x^2 (since adding 1 is negligible as xx becomes very large).

Thus, the expression becomes approximately:

xx2=1x0asx\frac{x}{x^2} = \frac{1}{x} \to 0 \quad \text{as} \quad x \to -\infty

So the limit is:

0\boxed{0}


b) limxx2x2+1\lim_{x \to -\infty} \frac{x^2}{x^2 + 1}

As xx \to -\infty:

  • The numerator x2x^2 grows very large.
  • The denominator x2+1x2x^2 + 1 \approx x^2 (since 1 becomes negligible).

Thus, the expression becomes approximately:

x2x2+1x2x2=1\frac{x^2}{x^2 + 1} \approx \frac{x^2}{x^2} = 1

So the limit is:

1\boxed{1}


c) limxexx\lim_{x \to -\infty} \frac{e^x}{x}

As xx \to -\infty:

  • The exponential function exe^x approaches 0 very quickly.
  • The linear term xx approaches -\infty.

Thus, we are dividing a very small number by a very large negative number, so the expression tends to:

00

So the limit is:

0\boxed{0}


d) limxarctan(x)ex\lim_{x \to \infty} \arctan(-x) e^x

As xx \to \infty:

  • arctan(x)π2\arctan(-x) \to -\frac{\pi}{2}.
  • exe^x grows exponentially fast to \infty.

Thus, the product is:

π2=-\frac{\pi}{2} \cdot \infty = -\infty

So the limit is:

\boxed{-\infty}


e) limxex/2\lim_{x \to \infty} e^{x/2}

As xx \to \infty, the expression ex/2e^{x/2} grows without bound since it is an exponential function. Therefore:

ex/2e^{x/2} \to \infty

So the limit is:

\boxed{\infty}


f) limxe2xx\lim_{x \to \infty} \frac{e^{2x}}{x}

As xx \to \infty:

  • e2xe^{2x} grows exponentially fast.
  • xx grows linearly.

Thus, the exponential growth dominates, and the expression tends to:

x=\frac{\infty}{x} = \infty

So the limit is:

\boxed{\infty}


Would you like further details on any of these solutions? Here are five related questions you could explore:

  1. What is the behavior of exponential functions as x±x \to \pm\infty?
  2. How can we use L'Hopital's Rule to evaluate indeterminate forms?
  3. Why do terms like x+1x + 1 become negligible for large values of xx?
  4. What is the significance of the arctan(x)\arctan(x) function's limits?
  5. How can the growth rates of different functions (e.g., exponential vs. polynomial) be compared?

Tip: For large xx, the term that grows fastest often dictates the behavior of the whole expression.

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Math Problem Analysis

Mathematical Concepts

Limits
Infinity
Exponential Functions
Arctangent

Formulas

lim (x -> ∞) f(x) = L
lim (x -> -∞) f(x) = L
Basic limits: lim (x -> ∞) 1/x = 0, lim (x -> ∞) e^x = ∞
Arctangent limit: lim (x -> ∞) arctan(x) = π/2

Theorems

L'Hopital's Rule
Limit Laws
Exponential Growth vs Polynomial Growth
Behavior of the arctan function as x approaches infinity

Suitable Grade Level

Grades 11-12 (Calculus)