Standard Limits and Expansions

1. Standard Limits

While L'Hôpital's rule is a powerful tool, it's often faster and more efficient to use standard, pre-proven limits. Committing these to memory is crucial for competitive exams.

A. Algebraic Limits

1. Polynomial and Rational Functions: If $f(x)$ is a polynomial or rational function and 'a' is in the domain of $f$, then $\lim_{x \to a} f(x) = f(a)$.
2. Power Formula: $$\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$$ where $n$ is any rational number.

B. Trigonometric Limits

1. $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$
2. $$\lim_{x \to 0} \frac{\tan x}{x} = 1$$
3. $$\lim_{x \to 0} \cos x = 1$$
4. $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$
5. $$\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$$
6. $$\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1$$ (Note: For limits involving trigonometric functions, the variable $x$ is assumed to be in radians.)

C. Exponential and Logarithmic Limits

1. $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$
2. $$\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a \quad (a > 0)$$
3. $$\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$$
4. $$\lim_{x \to \infty} (1 + \frac{1}{x})^x = e$$
5. $$\lim_{x \to 0} (1 + x)^{1/x} = e$$
6. $$\lim_{x \to \infty} (1 + \frac{a}{x})^x = e^a$$

D. The $1^{\infty}$ Form

If a limit is of the form $\lim_{x \to a} [f(x)]^{g(x)}$ and it results in the indeterminate form $1^{\infty}$ (i.e., $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$), it can be evaluated using the formula:

$$\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} g(x)[f(x)-1]}$$

2. Method of Series Expansion

For complex limits, especially when $x \to 0$, expanding functions into their Maclaurin series can simplify the problem significantly by allowing us to approximate the function with a polynomial.

Key Maclaurin Series Expansions (for $x$ near 0):

• $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
• $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \quad (-1 < x \leq 1)$
• $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
• $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
• $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$
• $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \quad (|x| < 1)$

Strategy: Replace the function with its series expansion, cancel out terms, and then evaluate the limit.

Solved Examples

Example 1: Using Trigonometric Limits

Question: Evaluate $\lim_{x \to 0} \frac{\sin(4x)}{\tan(7x)}$.

Solution: We can rewrite the expression to use the standard limits $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and $\lim_{u \to 0} \frac{\tan u}{u} = 1$. $$\lim_{x \to 0} \frac{\sin(4x)}{\tan(7x)} = \lim_{x \to 0} \left( \frac{\sin(4x)}{4x} \cdot 4x \right) \cdot \left( \frac{7x}{\tan(7x)} \cdot \frac{1}{7x} \right)$$ Rearranging the terms: $$= \lim_{x \to 0} \left( \frac{\sin(4x)}{4x} \right) \cdot \left( \frac{7x}{\tan(7x)} \right) \cdot \frac{4x}{7x}$$ As $x \to 0$, $4x \to 0$ and $7x \to 0$. So, the limits of the first two parts are 1. $$= (1) \cdot (1) \cdot \frac{4}{7} = \frac{4}{7}$$

Example 2: Using Exponential and Logarithmic Limits

Question: Evaluate $\lim_{x \to 0} \frac{e^{3x} - 1}{\ln(1+5x)}$.

Solution: This is a $\frac{0}{0}$ form. We can use the standard limits $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$ and $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$. Divide the numerator and denominator by $x$: $$\lim_{x \to 0} \frac{(e^{3x} - 1)/x}{(\ln(1+5x))/x} = \frac{\lim_{x \to 0} \frac{e^{3x} - 1}{x}}{\lim_{x \to 0} \frac{\ln(1+5x)}{x}}$$ To match the standard form, we adjust the constants: $$= \frac{\lim_{x \to 0} 3 \cdot \frac{e^{3x} - 1}{3x}}{\lim_{x \to 0} 5 \cdot \frac{\ln(1+5x)}{5x}} = \frac{3 \cdot (1)}{5 \cdot (1)} = \frac{3}{5}$$

Example 3: Using the $1^\infty$ Form

Question: Evaluate $\lim_{x \to \infty} \left( \frac{x+6}{x+1} \right)^{x+4}$.

Solution: This limit is of the form $1^\infty$. We can manipulate it to fit the standard form $\lim_{x \to \infty} (1 + \frac{a}{x})^x = e^a$. First, rewrite the base: $$\frac{x+6}{x+1} = \frac{(x+1)+5}{x+1} = 1 + \frac{5}{x+1}$$ So the limit becomes: $$\lim_{x \to \infty} \left( 1 + \frac{5}{x+1} \right)^{x+4} = \lim_{x \to \infty} \left( 1 + \frac{5}{x+1} \right)^{x+1} \cdot \left( 1 + \frac{5}{x+1} \right)^{3}$$ As $x \to \infty$, let $y = x+1$, so $y \to \infty$. The first part becomes $\lim_{y \to \infty} (1 + \frac{5}{y})^y = e^5$. The second part becomes $\lim_{x \to \infty} (1 + \frac{5}{x+1})^3 = (1+0)^3 = 1$. Therefore, the final limit is $e^5 \cdot 1 = e^5$.

Example 4: Using Series Expansion

Question: Evaluate $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^2 \sin x}$.

Solution: Direct substitution gives $\frac{0}{0}$. L'Hôpital's rule would be lengthy. Let's use series expansions up to $x^3$ for the numerator and denominator.

• $\cos x = 1 - \frac{x^2}{2!} + O(x^4)$
• $\sin x = x - \frac{x^3}{3!} + O(x^5)$

Numerator: $x(1 - \frac{x^2}{2}) - (x - \frac{x^3}{6}) = x - \frac{x^3}{2} - x + \frac{x^3}{6} = -\frac{3x^3 - x^3}{6} = -\frac{2x^3}{6} = -\frac{x^3}{3}$

Denominator: $x^2 \sin x \approx x^2(x) = x^3$

Now, the limit is: $$\lim_{x \to 0} \frac{-x^3/3}{x^3} = -\frac{1}{3}$$