Limit of a Function
We say the limit of \( f(x) \) as \( x \) approaches \( k \) is \( L \) if we can make the values of \( f(x) \) arbitrarily close to \( L \) by taking \( x \) to be sufficiently close to \( k \) (from both the left and right sides), but not equal to \( k \).
\[ \displaystyle \lim_{x \to k} f(x) = L \]
Theorems on Limits
Theorem 1. The limit of the sum of two or more functions is equal to the sum of their limits.
\[ \displaystyle \lim_{x \to a} \left[ f(x) + g(x) \right] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \]
Theorem 2. The limit of the product of two or more functions is equal to the product of their limits.
\[ \displaystyle \lim_{x \to a} \left[ f(x) \times g(x) \right] = \lim_{x \to a} f(x) \times \lim_{x \to a} g(x) \]
Theorem 3. The limit of the quotient of two functions is equal to the quotient of their limits, provided that the limit of the denominator is not zero.
\[ \displaystyle \lim_{x \to a} \left[ \dfrac{f(x)}{g(x)} \right] = \dfrac{\displaystyle \lim_{x \to a} f(x)}{\displaystyle \lim_{x \to a} g(x)}, \qquad \text{provided } \displaystyle \lim_{x \to a} g(x) \ne 0 \]
Standard Limits
- \( \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1 \)
- \( \displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{x} = 0 \)
- \( \displaystyle \lim_{x \to 0} \dfrac{\tan x}{x} = 1 \)
- \( \displaystyle \lim_{x \to 0} \dfrac{\arcsin x}{x} = 1 \)
- \( \displaystyle \lim_{x \to 0} \dfrac{\ln (1 + x)}{x} = 1 \)
- \( \displaystyle \lim_{x \to 0} \dfrac{e^x - 1}{x} = 1 \)
- \( \displaystyle \lim_{x \to \infty} \left( 1 + \dfrac{k}{x} \right)^x = e^k \)
- \( \displaystyle \lim_{x \to a} \dfrac{x^n - a^n}{x - a} = na^{n - 1} \)
The Limit Does Not Exist (DNE)
For two-sided limit \( \displaystyle \lim_{x \to k} f(x) = L \) to exist, three conditions must be met:
- The left-hand limit must exist.
- The right-hand limit must exist.
- They must be equal.
Failure to meet any of these conditions results in \( \displaystyle \lim_{x \to k} f(x) = \text{DNE} \).
Limits Involving Infinity
The Infinity Rule: If \( f(x) \) approaches the same infinity from both sides as \( x \to a \), then the limit is an infinite limit. If the left-hand and right-hand limits are unequal, or approach different infinities, then the limit is \( \text{DNE} \).
- Linear Growth: \( \displaystyle \lim_{u \to \infty} cu = \infty \)
- Scalar Growth: \( \displaystyle \lim_{u \to \infty} \left( \dfrac{u}{c} \right) = \infty \)
- The Vertical Asymptote: \( \displaystyle \lim_{u \to 0} \left( \dfrac{c}{u} \right) = \pm \infty \)
- The Horizontal Asymptote: \( \displaystyle \lim_{u \to \infty} \left( \dfrac{c}{u} \right) = 0 \)
- Exponential Growth: \( \displaystyle \lim_{u \to \infty} e^u = \infty \)
- Exponential Decay: \( \displaystyle \lim_{u \to \infty} e^{-u} = 0 \)
- Logarithmic Growth: \( \displaystyle \lim_{u \to \infty} \ln(u) = \infty \)
- Logarithmic Boundary: \( \displaystyle \lim_{u \to 0^+} \ln(u) = -\infty \)
L’Hospital’s Rule (Indeterminate Forms)
For limits that yield \( 0/0 \) or \( \infty/\infty \), the limit of the quotient is equal to the limit of the quotient of their derivatives.
\[ \displaystyle \lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)} \]
Note: Ensure the expression is in a fractional form before applying the rule. If the first derivative still results in an indeterminate form, apply the rule again.
Warning: L’Hospital’s Rule is not the Quotient Rule. Do not use \( \dfrac{v du - u dv}{v^2} \). Simply differentiate the numerator (\( f' \)) and the denominator (\( g' \)) individually.
The CALC-Method
When evaluating \( \displaystyle \lim_{x \to k} f(x) \), use the **CALC** function of your calculator to test values of \( x \) approaching \( k \) from both sides. Test \( x = k \pm 10^{-n} \) (for \( n = 2, 3, \dots, 6 \)) to observe the numerical trend. If the results from both the left (\( k - \Delta \)) and right (\( k + \Delta \)) converge to a stable constant, that constant is your limit.
Pro-Tip: When using the CALC-Method on your calculator, if the result shows a very large number (e.g., \( >10^6 \)), the limit likely approaches \( \infty \). If it alternates signs between \( k+\Delta \) and \( k-\Delta \), the limit is DNE.
Examples
Example 1
Evaluate: \( \displaystyle \lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x} \)
Solution:
Direct substitution gives \( 0/0 \).
**Using Elementary Method 1**
Rationalize the numerator
\( \begin{align}
\require{cancel} \dfrac{\sqrt{x + 4} - 2}{x} & = \dfrac{\sqrt{x + 4} - 2}{x} \cdot \dfrac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} \\
& = \dfrac{\left(\sqrt{x + 4}\right)^2 - 2^2}{x \left( \sqrt{x + 4} + 2 \right)} = \dfrac{(x + \cancel{4}) - \cancel{4}}{x \left( \sqrt{x + 4} + 2 \right)} \\
& = \dfrac{\cancel{x}}{\cancel{x} \left( \sqrt{x + 4} + 2 \right)} = \dfrac{1}{\sqrt{x + 4} + 2}
\end{align} \)
Substitute \( x = 0 \)
\( \dfrac{1}{\sqrt{4} + 2} = \dfrac{1}{4} \)
Hence,
\( \displaystyle \color{green}\boxed{\lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x} = \dfrac{1}{4}} \)
**Using L'Hospital's Rule**
\( \begin{align}
\displaystyle \lim_{x \to 0} \dfrac{\sqrt{x + 4} - 2}{x} &= \lim_{x \to 0}\dfrac{\dfrac{1}{2 \sqrt{x + 4}}}{1} = \lim_{x \to 0}\dfrac{1}{2 \sqrt{x + 4}} \\
&= \dfrac{1}{2 \sqrt{0 + 4}} = \color{green}\boxed{\dfrac{1}{4}}
\end{align} \)
Example 2
Evaluate: \( \displaystyle \lim_{x \to 0} x^2 \sin \left( \dfrac{1}{x} \right) \)
**Solution:**
As \( x \to 0 \), \( \dfrac{1}{x} \to \infty \), so \( \sin \left( \dfrac{1}{x} \right) \) oscillates between \( -1 \) and \( 1 \). Thus,
\( -1 \le \sin\left(\dfrac{1}{x}\right) \le 1 \)
Since \( x^2 \ge 0 \), multiplying the inequality by \( x^2 \) gives
\( -x^2 \le x^2 \sin \left(\dfrac{1}{x} \right) \le x^2 \)
Now, evaluate the limits of the bounding functions:
\( \displaystyle \lim_{x \to 0} (-x^2) = 0 \) and \( \displaystyle \lim_{x \to 0} (x^2) = 0 \)
By the **Squeeze Theorem**, it follows that
\( \displaystyle \color{green}\boxed{\lim_{x \to 0} x^2 \sin \left( \dfrac{1}{x} \right) = 0} \)
The graph of \( y = x^2 \sin \left( \dfrac{1}{x} \right) \) near \( x = 0 \):
Although \( \sin (1/x) \) oscillates infinitely as \( x \to 0 \), the factor \( x^2 \) forces the product to approach \( 0 \).
Example 3
Evaluate the \( \displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} \)
**Solution:**
By direct substitution, \( \dfrac{1 - \cos 0}{0^2} = \dfrac{0}{0} \).
Use L'Hospitals Rule:
\( \displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \lim_{x \to 0} \dfrac{\sin x}{2x} \)
which is still \( 0/0 \) when you substitute \( x = 0 \).
Use L'Hospitals Rule again:
\( \displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \lim_{x \to 0} \dfrac{\cos x}{2} \)
Substitute \( x = 0 \):
\( \dfrac{\cos x}{2} = \dfrac{1}{2} \)
Hence,
\( \displaystyle \color{green}\boxed{\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}} \)
For visual verification, the behavior of the function in the neighborhood of \( x = 0 \) is plotted below.
Example 4
Evaluate: \( \displaystyle \lim_{x \to 2} \dfrac{x^2-4}{x-2} \)
**Solution:**
Simplify the function:
\( \require{cancel} \dfrac{x^2-4}{x-2} = \dfrac{(\cancel{x - 2})(x + 2)}{\cancel{x - 2}} = x + 2 \)
Substitute \( x = 2 \):
\( x + 2 = 4 \)
Hence,
\( \displaystyle \color{green}\boxed{\lim_{x \to 2} \dfrac{x^2-4}{x-2} = 4} \)
Notice that the function is **undefined** at \( x = 2 \), as represented by the **open circle** in the graph below. In this case, although the function value does not exist at that specific point, the limit **does exist** because both paths converge to the same value.
Example 5
Evaluate: \( \displaystyle \lim_{x \to 2} f(x) \) where \( f(x) = \begin{cases} \dfrac{x^2-4}{x-2} & x \lt 2 \ 5 & x \ge 2 \end{cases} \)
**Solution:**
**Left-hand limit:**
Simplify \( \dfrac{x^2-4}{x-2} = \dfrac{(x - 2)(x + 2)}{x-2} = x+2 \)
\( \displaystyle \lim_{x \to 2^{-}} (x+2) = 4 \)
**Right-hand limit:**
\( f(x) = 5 \) for all \( x \gt 2 \).
\( \displaystyle \lim_{x \to 2^{+}} 5 = 5 \)
Because the left-hand limit is not equal to the right-hand limit, the limit does not exist.
\( \displaystyle \color{green}\boxed{\lim_{x \to 2} f(x) = \text{DNE}} \)
Although the function is **defined** at \( x = 2 \) (where \( f(2) = 5 \)), the limit **does not exist**. As shown in the graph below, the left-hand and right-hand paths approach different \( y \)-values, resulting in a jump discontinuity.
Example 6
Evaluate: \( \displaystyle \lim_{x \to 2} \dfrac{1}{x - 2} \)
**Solution:**
RHL = right-hand limit and LHL = left-hand limit
RHL: \( \displaystyle \lim_{x \to 2^{+}} \dfrac{1}{x - 2} = +\infty \)
LHL: \( \displaystyle \lim_{x \to 2^{-}} \dfrac{1}{x - 2} = -\infty \)
Hence,
\( \displaystyle \color{green}\boxed{\lim_{x \to 2} \dfrac{1}{x - 2} = \text{DNE}} \)
The graph goes downward to \( -\infty \) on the left and upward to \( +\infty \) on the right.
Example 7
Evaluate: \( \displaystyle \lim_{x \to 2} \dfrac{1}{(x - 2)^2} \)
**Solution:**
RHL: \( \displaystyle \lim_{x \to 2^{+}} \dfrac{1}{(x - 2)^2} = +\infty \)
LHL: \( \displaystyle \lim_{x \to 2^{-}} \dfrac{1}{(x - 2)^2} = +\infty \)
Hence,
\( \displaystyle \color{green}\boxed{\lim_{x \to 2} \dfrac{1}{(x - 2)^2} = +\infty} \)
The graph rises to \( +\infty \) on both sides of \( x = 2 \).
Example 8
Find the horizontal asymptote of the function \( f(x) = \dfrac{3x^2 - 5}{x^2 + 1} \).
**Solution:**
To find the horizontal asymptote, take \( \displaystyle \lim_{x \to \infty} f(x) \).
\( \begin{align}
\displaystyle \lim_{x \to \infty} f(x) &= \lim_{x \to \infty} \dfrac{3x^2 - 5}{x^2 + 1} = \lim_{x \to \infty} \dfrac{\dfrac{3x^2}{x^2} - \dfrac{5}{x^2}}{\dfrac{x^2}{x^2} + \dfrac{1}{x^2}} \\
&= \lim_{x \to \infty} \dfrac{3 - \dfrac{5}{x^2}}{1 + \dfrac{1}{x^2}} = \dfrac{3 - \dfrac{5}{\infty^2}}{1 + \dfrac{1}{\infty^2}} \\
&= \dfrac{3 - 0}{1 + 0} = 3
\end{align} \)
Hence, the horizontal asymptote is
\( \color{green}\boxed{y = 3} \)
It is important to note that the answer for an asymptote is an equation of a line (\( y = 3 \)), not just a number (3).
If you want, I can also clean this further by removing the image placeholders and extra spacing.