20 Limits at Infinity (Detailed Explanations)

Limits at Infinity

20 Detailed Exercises (Form $\infty/\infty$)

Core Concept: To solve limits of the form $\infty/\infty$, divide both the numerator and the denominator by the highest power of $x$ present in the denominator.

Exercise 1 Basic

$$ \lim_{x \to \infty} \dfrac{3x + 5}{2x - 1} $$

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Step 1: Identify the highest power of $x$ in the denominator. Here, it is $x^1$ (or simply $x$).

Step 2: Divide every term in the numerator and denominator by $x$.

$$ \lim_{x \to \infty} \frac{\frac{3x}{x} + \frac{5}{x}}{\frac{2x}{x} - \frac{1}{x}} = \lim_{x \to \infty} \frac{3 + \frac{5}{x}}{2 - \frac{1}{x}} $$

Step 3: Apply the limit. As $x \to \infty$, terms like $\frac{5}{x}$ approach 0.

$$ = \frac{3 + 0}{2 - 0} = \mathbf{\frac{3}{2}} $$

Exercise 2 Basic

$$ \lim_{x \to \infty} \frac{4x^2 - x}{x^2 + 3} $$

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Step 1: The highest power in the denominator is $x^2$.

Step 2: Divide every term by $x^2$.

$$ \lim_{x \to \infty} \frac{\frac{4x^2}{x^2} - \frac{x}{x^2}}{\frac{x^2}{x^2} + \frac{3}{x^2}} = \lim_{x \to \infty} \frac{4 - \frac{1}{x}}{1 + \frac{3}{x^2}} $$

Step 3: As $x \to \infty$, $\frac{1}{x} \to 0$ and $\frac{3}{x^2} \to 0$.

$$ = \frac{4 - 0}{1 + 0} = \mathbf{4} $$

Exercise 3 Basic

$$ \lim_{x \to \infty} \frac{x}{x^2 - 1} $$

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Step 1: Highest power in denominator is $x^2$.

Step 2: Divide everything by $x^2$.

$$ \lim_{x \to \infty} \frac{\frac{x}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1 - \frac{1}{x^2}} $$

Step 3: Apply limit.

$$ = \frac{0}{1 - 0} = \mathbf{0} $$

Note: If degree of denominator > degree of numerator, limit is 0.

Exercise 4 Basic

$$ \lim_{x \to \infty} \frac{x^3 + 1}{2x^2} $$

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Step 1: Highest power in denominator is $x^2$.

Step 2: Divide everything by $x^2$.

$$ \lim_{x \to \infty} \frac{\frac{x^3}{x^2} + \frac{1}{x^2}}{\frac{2x^2}{x^2}} = \lim_{x \to \infty} \frac{x + \frac{1}{x^2}}{2} $$

Step 3: As $x \to \infty$, the numerator goes to $\infty$.

$$ = \frac{\infty + 0}{2} = \mathbf{\infty} $$

Exercise 5 Medium

$$ \lim_{x \to \infty} \frac{5x^3 - 2x^2 + 1}{2x^3 + 4} $$

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Observation: Both numerator and denominator have the same degree (3).

Shortcut: The limit is simply the ratio of the leading coefficients.

$$ = \frac{5}{2} = \mathbf{2.5} $$

Detailed: Divide by $x^3$: $\frac{5 - 2/x + 1/x^3}{2 + 4/x^3} \to \frac{5}{2}$.

Exercise 6 Medium

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

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Step 1: Expand the numerator.

$$ (2x+1)^2 = 4x^2 + 4x + 1 $$

Step 2: Rewrite the limit.

$$ \lim_{x \to \infty} \frac{4x^2 + 4x + 1}{x^2 - 1} $$

Step 3: Divide by $x^2$.

$$ = \frac{4}{1} = \mathbf{4} $$

Exercise 7 Medium

$$ \lim_{x \to \infty} \frac{x^2}{3x^2 + x - 5} $$

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Step 1: Identify highest degree ($x^2$).

Step 2: Divide all by $x^2$.

$$ \lim_{x \to \infty} \frac{1}{3 + \frac{1}{x} - \frac{5}{x^2}} $$

Step 3: Evaluate.

$$ = \frac{1}{3 + 0 - 0} = \mathbf{\frac{1}{3}} $$

Exercise 8 Hard

$$ \lim_{x \to \infty} \frac{\sqrt{x^2 + 1}}{2x - 3} $$

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Step 1: The highest power in the denominator is $x$.

Step 2: Divide by $x$. Note that inside the square root, $x$ becomes $x^2$ (since $x = \sqrt{x^2}$ for $x>0$).

$$ \frac{\sqrt{\frac{x^2}{x^2} + \frac{1}{x^2}}}{\frac{2x}{x} - \frac{3}{x}} = \frac{\sqrt{1 + \frac{1}{x^2}}}{2 - \frac{3}{x}} $$

Step 3: Evaluate.

$$ = \frac{\sqrt{1 + 0}}{2 - 0} = \mathbf{\frac{1}{2}} $$

Exercise 9 Hard

$$ \lim_{x \to -\infty} \frac{\sqrt{x^2 + 4}}{3x + 1} $$

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Crucial Step: When $x \to -\infty$, $\sqrt{x^2} = |x| = -x$.

We divide the denominator by $x$, but the numerator by $-\sqrt{x^2}$ to account for the sign.

$$ \lim_{x \to -\infty} \frac{\sqrt{1 + \frac{4}{x^2}}}{3 + \frac{1}{x}} \times (-1) $$

The $(-1)$ comes from $x = -\sqrt{x^2}$.

$$ = -\frac{\sqrt{1}}{3} = \mathbf{-\frac{1}{3}} $$

Exercise 10 Medium

$$ \lim_{x \to \infty} \frac{1 - x^2}{x^3 - x + 1} $$

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Analysis: Numerator degree is 2. Denominator degree is 3.

Since degree(bottom) > degree(top), the limit is 0.

Proof: Divide by $x^3$.

$$ \frac{\frac{1}{x^3} - \frac{1}{x}}{1 - \frac{1}{x^2} + \frac{1}{x^3}} \to \frac{0}{1} = \mathbf{0} $$

Exercise 11 Hard

$$ \lim_{x \to \infty} \sqrt{x^2 + 2x} - x $$

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Problem: This is form $\infty - \infty$. We must rationalize.

Step 1: Multiply by conjugate $\frac{\sqrt{x^2+2x}+x}{\sqrt{x^2+2x}+x}$.

$$ \lim_{x \to \infty} \frac{(x^2+2x) - x^2}{\sqrt{x^2+2x}+x} = \lim_{x \to \infty} \frac{2x}{\sqrt{x^2+2x}+x} $$

Step 2: Divide by $x$.

$$ \frac{2}{\sqrt{1+2/x} + 1} = \frac{2}{1+1} = \mathbf{1} $$

Exercise 12 Hard

$$ \lim_{x \to \infty} (\sqrt{x^2 + x} - \sqrt{x^2 - x}) $$

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Step 1: Rationalize by multiplying by the conjugate sum.

$$ \frac{(x^2+x)-(x^2-x)}{\sqrt{x^2+x} + \sqrt{x^2-x}} = \frac{2x}{\sqrt{x^2+x} + \sqrt{x^2-x}} $$

Step 2: Divide by $x$.

$$ \frac{2}{\sqrt{1+1/x} + \sqrt{1-1/x}} = \frac{2}{1+1} = \mathbf{1} $$

Exercise 13 Medium

$$ \lim_{x \to \infty} \frac{(x+1)(2x-1)}{(3x+2)(x-5)} $$

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Method: We only need the leading terms.

Numerator leading term: $x \cdot 2x = 2x^2$.

Denominator leading term: $3x \cdot x = 3x^2$.

$$ \text{Limit} = \frac{2x^2}{3x^2} = \mathbf{\frac{2}{3}} $$

Exercise 14 Hard

$$ \lim_{x \to \infty} \frac{x}{\sqrt{x^2+1}} $$

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Method: Divide by $x$. Remember $\sqrt{x^2}=x$.

$$ \frac{x/x}{\sqrt{x^2/x^2 + 1/x^2}} = \frac{1}{\sqrt{1 + 1/x^2}} $$

$$ = \frac{1}{\sqrt{1+0}} = \mathbf{1} $$

Exercise 15 Hard

$$ \lim_{x \to -\infty} \frac{x}{\sqrt{x^2+1}} $$

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Crucial Difference: Here $x \to -\infty$, so $\sqrt{x^2} = -x$.

$$ \frac{x}{-\sqrt{x^2+1}} = \frac{x}{-|x|\sqrt{1+1/x^2}} = \frac{x}{-(-x)\sqrt{1+...}} $$

Dividing numerator $x$ by denominator's effective $-x$ gives $\mathbf{-1}$.

Exercise 16 Medium

$$ \lim_{x \to \infty} \frac{2x^3}{x^2 + 1} - 2x $$

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Step 1: Find a common denominator to combine terms.

$$ \frac{2x^3 - 2x(x^2+1)}{x^2+1} = \frac{2x^3 - 2x^3 - 2x}{x^2+1} $$

$$ = \frac{-2x}{x^2+1} $$

Step 2: Degree of bottom > degree of top. Limit is $\mathbf{0}$.

Exercise 17 Advanced

$$ \lim_{x \to \infty} \frac{3^x + 4^x}{3^x - 4^x} $$

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Method: Divide by the fastest growing term ($4^x$).

$$ \lim_{x \to \infty} \frac{(3/4)^x + 1}{(3/4)^x - 1} $$

Since $3/4 < 1$, $(3/4)^x \to 0$ as $x \to \infty$.

$$ = \frac{0 + 1}{0 - 1} = \mathbf{-1} $$

Exercise 18 Advanced

$$ \lim_{x \to \infty} \frac{\ln(x^2)}{\ln(x^3)} $$

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Method: Use log properties $\ln(a^b) = b \ln a$.

Numerator: $2 \ln x$. Denominator: $3 \ln x$.

$$ \lim_{x \to \infty} \frac{2 \ln x}{3 \ln x} = \mathbf{\frac{2}{3}} $$

Exercise 19 Medium

$$ \lim_{x \to \infty} \frac{5x^2}{\sqrt{9x^4 + 1}} $$

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Step 1: Highest effective power in denominator is $\sqrt{x^4} = x^2$.

Step 2: Divide top by $x^2$ and bottom by $\sqrt{x^4}$.

$$ \frac{5}{\sqrt{9 + 1/x^4}} = \frac{5}{\sqrt{9}} = \mathbf{\frac{5}{3}} $$

Exercise 20 Advanced

$$ \lim_{n \to \infty} \frac{1^2 + 2^2 + ... + n^2}{n^3} $$

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Step 1: Use the sum formula for squares: $\frac{n(n+1)(2n+1)}{6}$.

$$ \lim_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3} $$

Step 2: Expand leading terms: $n \cdot n \cdot 2n = 2n^3$.

$$ \lim_{n \to \infty} \frac{2n^3 + ...}{6n^3} = \frac{2}{6} = \mathbf{\frac{1}{3}} $$

~ Edu-IKH ~