✨ Math Magic: Exploring Limits ✨
Show Me the Math Magic!
1 Limit of (1 - cos x)/x² as x→0
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \]
Show Solution
Step 1: Direct substitution gives 0/0 (indeterminate form)
Step 2: Apply l'Hôpital's Rule by differentiating numerator and denominator:
\[ \frac{d}{dx}(1 - \cos x) = \sin x \]
\[ \frac{d}{dx}x^2 = 2x \]
Step 3: New limit:
\[ \lim_{x \to 0} \frac{\sin x}{2x} \]
Still 0/0, so apply l'Hôpital's Rule again
Step 4: Differentiate again:
\[ \frac{d}{dx}\sin x = \cos x \]
\[ \frac{d}{dx}2x = 2 \]
Step 5: Final limit:
\[ \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2} \]
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Fun Fact: This limit is important in physics for calculating small-angle approximations!
2 Limit of (2x² - 3)/(x² - 5x + 3) as x→∞
\[ \lim_{x \to \infty} \frac{2x^2 - 3}{x^2 - 5x + 3} \]
Show Solution
Step 1: For limits at infinity, divide numerator and denominator by the highest power of x (x²):
\[ \frac{2 - \frac{3}{x^2}}{1 - \frac{5}{x} + \frac{3}{x^2}} \]
Step 2: As x→∞, terms with x in denominator approach 0:
\[ \frac{2 - 0}{1 - 0 + 0} = 2 \]
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Fun Fact: This shows that for large x, the function behaves like 2x²/x² = 2!
3 Limit of x/log x as x→∞
\[ \lim_{x \to \infty} \frac{x}{\log x} \]
Show Solution
Step 1: This is ∞/∞ form, so apply l'Hôpital's Rule
Step 2: Differentiate numerator and denominator:
\[ \frac{d}{dx}x = 1 \]
\[ \frac{d}{dx}\log x = \frac{1}{x} \]
Step 3: New limit:
\[ \lim_{x \to \infty} \frac{1}{1/x} = \lim_{x \to \infty} x = \infty \]
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Fun Fact: This shows that x grows much faster than log x as x becomes large!
4 Limit of sec x/tan x as x→π/2
\[ \lim_{x \to \frac{\pi}{2}} \frac{\sec x}{\tan x} \]
Show Solution
Step 1: Rewrite in terms of sin and cos:
\[ \sec x = \frac{1}{\cos x}, \tan x = \frac{\sin x}{\cos x} \]
So:
\[ \frac{\sec x}{\tan x} = \frac{1/\cos x}{\sin x/\cos x} = \frac{1}{\sin x} = \csc x \]
Step 2: Now evaluate:
\[ \lim_{x \to \frac{\pi}{2}} \csc x = \csc \left(\frac{\pi}{2}\right) = 1 \]
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Fun Fact: This trigonometric identity simplifies the problem without needing l'Hôpital's Rule!
5 Limit of e⁻ˣ√x as x→∞
\[ \lim_{x \to \infty} e^{-x} \sqrt{x} \]
Show Solution
Step 1: Rewrite the expression:
\[ e^{-x}\sqrt{x} = \frac{\sqrt{x}}{e^x} \]
Step 2: This is ∞/∞ form, so apply l'Hôpital's Rule
Step 3: Differentiate numerator and denominator:
\[ \frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}} \]
\[ \frac{d}{dx}e^x = e^x \]
Step 4: New limit:
\[ \lim_{x \to \infty} \frac{1}{2\sqrt{x}e^x} = 0 \]
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Fun Fact: Exponential functions grow much faster than any power function!
6 Limit of (1/sin x - 1/x) as x→0
\[ \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right) \]
Show Solution
Step 1: Combine the fractions:
\[ \frac{x - \sin x}{x \sin x} \]
Step 2: Direct substitution gives 0/0, so apply l'Hôpital's Rule
Step 3: Differentiate numerator and denominator:
\[ \frac{1 - \cos x}{\sin x + x \cos x} \]
Still 0/0, so apply l'Hôpital's Rule again
Step 4: Differentiate again:
\[ \frac{\sin x}{2\cos x - x \sin x} \]
Step 5: Now evaluate at x→0:
\[ \frac{0}{2 - 0} = 0 \]
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Fun Fact: This shows that 1/sin x and 1/x approach each other as x→0!
7 Limit of [2/(x²-1) - x/(x-1)] as x→1⁻
\[ \lim_{x \to 1^-} \left( \frac{2}{x^2 - 1} - \frac{x}{x - 1} \right) \]
Show Solution
Step 1: Factor denominators and combine fractions:
\[ \frac{2}{(x-1)(x+1)} - \frac{x}{x-1} = \frac{2 - x(x+1)}{(x-1)(x+1)} \]
Step 2: Simplify numerator:
\[ 2 - x^2 - x = -x^2 - x + 2 \]
Step 3: Factor numerator:
\[ -(x^2 + x - 2) = -(x+2)(x-1) \]
Step 4: Cancel (x-1) terms:
\[ \frac{-(x+2)}{x+1} \]
Step 5: Now evaluate as x→1⁻:
\[ \frac{-(1+2)}{1+1} = -\frac{3}{2} \]
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Fun Fact: Combining fractions often simplifies complex limits!
8 Limit of xˣ as x→0⁺
\[ \lim_{x \to 0^+} x^x \]
Show Solution
Step 1: Let y = xˣ, then take natural log:
\[ \ln y = x \ln x \]
Step 2: Find limit of ln y:
\[ \lim_{x \to 0^+} x \ln x \]
This is 0·(-∞), rewrite as ln x / (1/x)
Step 3: Apply l'Hôpital's Rule:
\[ \frac{d}{dx}\ln x = \frac{1}{x} \]
\[ \frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2} \]
So:
\[ \frac{1/x}{-1/x^2} = -x \]
Step 4: Now:
\[ \lim_{x \to 0^+} -x = 0 \]
So:
\[ \lim_{x \to 0^+} \ln y = 0 \]
Step 5: Therefore:
\[ \lim_{x \to 0^+} y = e^0 = 1 \]
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Fun Fact: 0⁰ is indeterminate, but the limit of xˣ as x→0⁺ is 1!
9 Limit of (1 + 1/x)ˣ as x→∞
\[ \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^x \]
Show Solution
Step 1: Let y = (1 + 1/x)ˣ, take natural log:
\[ \ln y = x \ln(1 + 1/x) \]
Step 2: Rewrite as:
\[ \frac{\ln(1 + 1/x)}{1/x} \]
As x→∞, this is 0/0 form
Step 3: Apply l'Hôpital's Rule:
\[ \frac{d}{dx}\ln(1 + 1/x) = \frac{-1/x^2}{1 + 1/x} \]
\[ \frac{d}{dx}1/x = -1/x^2 \]
Step 4: Simplify:
\[ \frac{-1/x^2}{1 + 1/x} ÷ \frac{-1/x^2}{1} = \frac{1}{1 + 1/x} \]
Step 5: Now as x→∞:
\[ \frac{1}{1 + 0} = 1 \]
So:
\[ \lim_{x \to \infty} \ln y = 1 \]
Step 6: Therefore:
\[ \lim_{x \to \infty} y = e^1 = e \]
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Fun Fact: This is the famous definition of the number e (≈2.71828)!
10 Limit of (sin x)^(tan x) as x→π/2
\[ \lim_{x \to \frac{\pi}{2}} (\sin x)^{\tan x} \]
Show Solution
Step 1: Let y = (sin x)^(tan x), take natural log:
\[ \ln y = \tan x \cdot \ln(\sin x) \]
Step 2: Rewrite as:
\[ \frac{\ln(\sin x)}{\cot x} \]
As x→π/2, this is 0/0 form
Step 3: Apply l'Hôpital's Rule:
\[ \frac{d}{dx}\ln(\sin x) = \frac{\cos x}{\sin x} \]
\[ \frac{d}{dx}\cot x = -\csc^2 x \]
Step 4: Simplify:
\[ \frac{\cos x/\sin x}{-1/\sin^2 x} = -\cos x \sin x \]
Step 5: Now as x→π/2:
\[ -\cos(\pi/2)\sin(\pi/2) = -0 \cdot 1 = 0 \]
So:
\[ \lim_{x \to \pi/2} \ln y = 0 \]
Step 6: Therefore:
\[ \lim_{x \to \pi/2} y = e^0 = 1 \]
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Fun Fact: Even though sin(π/2)=1, the exponent tan(π/2) is undefined, but the limit exists!
11 Limit of (cos x)^(1/x²) as x→0⁺
\[ \lim_{x \to 0^+} (\cos x)^{\frac{1}{x^2}} \]
Show Solution
Step 1: Let y = (cos x)^(1/x²), take natural log:
\[ \ln y = \frac{\ln(\cos x)}{x^2} \]
Step 2: As x→0⁺, this is 0/0 form, apply l'Hôpital's Rule
Step 3: Differentiate numerator and denominator:
\[ \frac{d}{dx}\ln(\cos x) = \frac{-\sin x}{\cos x} \]
\[ \frac{d}{dx}x^2 = 2x \]
Step 4: New limit:
\[ \lim_{x \to 0^+} \frac{-\sin x}{2x \cos x} \]
Still 0/0, apply l'Hôpital's Rule again
Step 5: Differentiate again:
\[ \frac{-\cos x}{2\cos x - 2x \sin x} \]
Step 6: Now evaluate at x→0⁺:
\[ \frac{-1}{2 - 0} = -\frac{1}{2} \]
So:
\[ \lim_{x \to 0^+} \ln y = -\frac{1}{2} \]
Step 7: Therefore:
\[ \lim_{x \to 0^+} y = e^{-1/2} = \frac{1}{\sqrt{e}} \]
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Fun Fact: This limit shows how quickly cos x approaches 1 as x→0!
12 Continuous Compound Interest
Show that as n→∞, A = A₀(1 + r/n)^(nt) → A₀e^(rt)
Show Solution
Step 1: Rewrite the expression:
\[ A = A_0 \left( 1 + \frac{r}{n} \right)^{nt} = A_0 \left[ \left( 1 + \frac{r}{n} \right)^{n/r} \right]^{rt} \]
Step 2: Let m = n/r, so n = mr:
\[ A = A_0 \left[ \left( 1 + \frac{1}{m} \right)^m \right]^{rt} \]
Step 3: As n→∞, m→∞:
\[ \lim_{m \to \infty} \left( 1 + \frac{1}{m} \right)^m = e \]
Step 4: Therefore:
\[ A = A_0 e^{rt} \]
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Fun Fact: This is why banks use continuous compounding - it's the maximum possible growth!