Glossary

Before finding the limit of $\frac{x^2 - 4}{x - 2}$ as $x o 2$, you would use algebraic manipulation to factor the numerator and cancel the common term.

If $f(x) = x^2$ and $g(x) = \sin x$, then $g(f(x)) = \sin(x^2)$ is a composite function.

To rationalize the denominator of $\frac{1}{\sqrt{3}-1}$, you would multiply by its conjugate, $\sqrt{3}+1$, on both the numerator and denominator.

The function $f(x) = x^2$ is a continuous function because its graph is a smooth, unbroken parabola.

A polynomial function like $f(x) = x^2 + 3x - 1$ is a continuous function for all real numbers, as its graph is a smooth, unbroken curve.

The expression $x^2 - 5x + 6$ can be factorized into $(x-2)(x-3)$.

If $\lim_{x o \infty} f(x) = 7$, then 7 is the finite limit at infinity, which directly corresponds to a horizontal asymptote at $y=7$.

Observing the graph of $f(x) = \frac{1}{x}$ clearly shows its behavior near $x=0$ and as $x$ extends towards positive or negative infinity.

The line y = 3 is a horizontal asymptote for the function $k(x) = (3x^2+1)/(x^2+x)$, meaning the function's graph flattens out towards y = 3 as x gets very large or very small.

The function $f(x) = \frac{2x+1}{x-3}$ has a horizontal asymptote at $y=2$, indicating that as $x$ becomes very large or very small, the function's values get arbitrarily close to 2.

When trying to find $\lim_{x\to 0} \frac{\sin x}{x}$, direct substitution yields the indeterminate form $\frac{0}{0}$, requiring L'Hopital's Rule or series expansion.

As x approaches 0, the function $f(x) = 1/x^2$ demonstrates an infinite limit, shooting up towards positive infinity.

The function $f(x) = 1/x$ has infinite limits as $x$ approaches 0, tending to $\infty$ from the right and $-\infty$ from the left.

If $\lim_{x \to 0} f(x) = 5$ and $\lim_{x \to 0} g(x) = 0$, the infinite limits of quotients property helps determine if $\lim_{x \to 0} \frac{f(x)}{g(x)}$ is $\infty$ or $-\infty$ based on the sign of $g(x)$.

When analyzing $f(x) = |x|/x$, the left-hand limit as $x \to 0^-$ is -1.

The limit of $f(x) = \frac{x^2 - 9}{x - 3}$ as $x$ approaches 3 is 6, even though the function is undefined at $x=3$.

For the function $f(x) = x+2$, the limit as $x$ approaches 3 is 5, meaning the function's output gets arbitrarily close to 5 as the input gets close to 3.

For the function $f(x) = x^2$, the limit as x approaches 3 is 9, meaning as x gets infinitesimally close to 3, $f(x)$ gets arbitrarily close to 9.

For the function $h(x) = (2x+1)/x$, the limit at infinity as x approaches positive infinity is 2, showing the function stabilizes at that value.

For $f(x) = \frac{1}{x}$, the limit at infinity as $x \to \infty$ is 0, indicating the function approaches the x-axis.

When analyzing the behavior of $y = 1/x^2$ near $x=0$, the limit of 1/x^n as x approaches 0 tells us it tends to $\infty$ from both sides because $n=2$ is even.

For $\lim_{x \to 0} \sin(x^2)$, since $\sin(x)$ is continuous, you can find $\lim_{x \to 0} x^2 = 0$ and then evaluate $\sin(0)$, applying the limit of a composite function.

If you're calculating the speed of light, which is a constant, its limit of a constant function as time approaches any point is still the speed of light.

If $\lim_{x \to 1} f(x) = 5$, then $\lim_{x \to 1} (4f(x))$ uses the limit of a multiple of a function to become $4 \times 5 = 20$.

When evaluating $\lim_{x \to 3} (x \cdot \sin x)$, you can use the limit of a product of functions to multiply $\lim_{x \to 3} x$ by $\lim_{x \to 3} \sin x$.

To find $\lim_{x \to 1} \frac{x^2+1}{x+1}$, you apply the limit of a quotient of functions, calculating the limit of the numerator and denominator separately.

To find $\lim_{x \to 0} (x^2 + \cos x)$, you can apply the limit of a sum or difference of functions to evaluate $\lim_{x \to 0} x^2$ and $\lim_{x \to 0} \cos x$ separately.

If you need to find $\lim_{x \to 2} (x^3)^2$, you can first find $\lim_{x \to 2} x^3$ and then square the result, using the limit of the power of a function.

Using limit properties, we can break down $\lim_{x \to 2} (3x^2 - 4x + 5)$ into simpler parts like $\lim_{x \to 2} (3x^2)$ and $\lim_{x \to 2} (4x)$.

When analyzing the function $$f(x) = \frac{\sin(x)}{x}$$, the limit as x approaches 0 is 1, even though the function itself is undefined at x=0.

The function $$f(x) = \tan(x)$$ has a nonexistent limit as x approaches $$\pi/2$$ because the function becomes unbounded.

For a piecewise function, checking the one-sided limit as $x \to 0^+$ for $f(x) = x^2$ would give 0, while $x \to 0^-$ for $f(x) = x-1$ would give -1.

For a piecewise function like $$f(x) = |x|/x$$, the one-sided limit as x approaches 0 from the left is -1, and from the right is 1.

For $f(x) = 1/x$, the one-sided limit as x approaches 0 from the right ($x o 0^+$) is positive infinity, while from the left ($x o 0^-$) it's negative infinity.

The function $$f(x) = \sin(1/x)$$ oscillates infinitely many times between -1 and 1 as x approaches 0, preventing a limit from existing.

A common example of a piecewise function is $f(x) = \begin{cases} x+1, & x < 0 \\ x^2, & x \geq 0 \end{cases}$, which behaves differently on either side of $x=0$.

In the expression $\frac{x^2 - 4}{x - 2}$, the entire fraction represents a quotient of two polynomial expressions.

The function $f(x) = \frac{x^2+1}{x-4}$ is a quotient function, and its denominator becoming zero at $x=4$ suggests a potential vertical asymptote.

The reciprocal of $x^2$ is $\frac{1}{x^2}$, which is useful when finding limits at infinity.

For the function $f(x) = |x|/x$, the right-hand limit as $x \to 0^+$ is 1.

To find $\lim_{x\to 2} \frac{x^2-4}{x-2}$, you can use simplifying by factoring the numerator to $(x-2)(x+2)$, canceling $(x-2)$, and then substituting into $x+2$ to get 4.

If you can show that $g(x) \le f(x) \le h(x)$ and both $g(x)$ and $h(x)$ approach 5 as $x$ approaches 2, then by the Squeeze Theorem, $f(x)$ must also approach 5.

To find the limit of $f(x) = 2x + 5$ as $x$ approaches 3, you can use substitution to get $2(3) + 5 = 11$.

Numbers like $\sqrt{2}$, $\sqrt{7}$, or $\sqrt{12}$ (which simplifies to $2\sqrt{3}$) are examples of surds.

By examining a table of values for $f(x) = \frac{\sin x}{x}$ as $x$ gets closer to 0, you can estimate that its limit is 1.

To simplify $\frac{1 - \cos^2 x}{x^2}$ for a limit problem, you would use the trigonometric identity $\sin^2 x + \cos^2 x = 1$ to replace $1 - \cos^2 x$ with $\sin^2 x$.

When faced with $\lim_{x o 0} \frac{1 - \cos 4x}{x}$, you can rearrange it to utilize this specific Trigonometric Limit Theorem.

To find $\lim_{x o 0} \frac{\sin 5x}{2x}$, you can manipulate the expression to apply the Trigonometric Limit Theorem for $\frac{\sin x}{x}$.

If a function smoothly approaches 5 as x gets closer to 2 from both directions, then its two-sided limit at x=2 is 5.

The function $$f(x) = 1/x^2$$ is unbounded as x approaches 0, shooting up towards positive infinity.

As $x$ approaches 0, the function $f(x) = \frac{1}{x^2}$ becomes unbounded, meaning its values shoot up towards positive infinity.

The line x = 2 is a vertical asymptote for the function $g(x) = 1/(x-2)$, indicating the function's values become unbounded near x = 2.

The function $f(x) = \frac{1}{x-5}$ has a vertical asymptote at $x=5$, because the function's values become infinitely large or small as $x$ gets closer to 5.