Q.Show that: $\displaystyle\lim_{x\to 0^+} x\left[\left\lfloor\dfrac{1}{x}\right\rfloor+\left\lfloor\dfrac{2}{x}\right\rfloor+\ldots+\left\lfloor\dfrac{15}{x}\right\rfloor\right]=120$ OR There are two identical urns containing respectively, 6 black and 4 red balls, 2 black and 2 red balls. An urn is chosen at random and a ball is drawn from it.

From Guessing to Computing

Tables and graphs are fine for building intuition, but they are impractical as a general method — you cannot tabulate your way through an exam. This concept collects the theorems that let you compute a limit algebraically, in a "mechanical" way, without ever touching a table.

Direct substitution for polynomials

Theorem 9.1. If $P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$ is a polynomial, then for any real $x_0$,

In other words: for a polynomial, the limit is just direct substitution. This works because a polynomial has no "trouble points" — it's defined and smooth everywhere, so there's nothing for a limit to disagree with. (A constant function is the degree-0 special case: $\lim_{x\to x_0} c = c$.)

The algebra of limits

Theorem 9.2. Suppose $\lim_{x\to x_0} f(x)$ and $\lim_{x\to x_0} g(x)$ both exist, and $c$ is a constant. Then all of the following exist, and:

• Constant multiple: $\displaystyle\lim_{x\to x_0} c\,f(x) = c \lim_{x\to x_0} f(x)$
• Sum/Difference: $\displaystyle\lim_{x\to x_0} [f(x) \pm g(x)] = \lim_{x\to x_0} f(x) \pm \lim_{x\to x_0} g(x)$
• Product: $\displaystyle\lim_{x\to x_0} [f(x)\,g(x)] = \lim_{x\to x_0} f(x) \cdot \lim_{x\to x_0} g(x)$
• Quotient: $\displaystyle\lim_{x\to x_0} \frac{f(x)}{g(x)} = \frac{\lim_{x\to x_0} f(x)}{\lim_{x\to x_0} g(x)}$, provided $\lim_{x\to x_0} g(x) \neq 0$.

These rules extend to any finite number of functions being added, multiplied, etc.

The power rule

Theorem 9.3. If $\lim_{x\to x_0} f(x)$ exists, then for any positive integer (in fact any rational) $n$,

This is really just the product rule applied $n$ times to $f\cdot f \cdots f$.

The difference-of-powers limit

Theorem 9.4. For a positive integer $n$ (and in fact for any rational exponent),

Why it's true, in brief: factor $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \cdots + a^{n-1})$. Cancel the common factor $(x-a)$ (legitimate since we only ever consider $x \neq a$ when taking the limit) to leave $n$ terms, each of which tends to $a^{n-1}$ as $x \to a$ — giving a total of $n\,a^{n-1}$. This single result is a shortcut generator: it turns an entire family of "$0/0$"-looking quotients into instant answers, and — though its proof waits for a later chapter — it is secretly the seed of the derivative of $x^n$.

The three evaluation techniques for a $0/0$ form at a finite point

When direct substitution into a quotient produces $\frac{0}{0}$, the quotient theorem is blocked (the denominator's limit is $0$), so you cannot stop — you must first rewrite the expression so the offending factor cancels.

1. Factor and cancel. If both numerator and denominator vanish at $x_0$, then $(x - x_0)$ is a common factor of both — pull it out and cancel. …