Limits and Continuity

Example

Find $\displaystyle\lim_{x \to 0} \frac{|x|}{x}$.

Since the one-sided limits are not equal, the limit does not exist.

Example

Find $\displaystyle\lim_{x \to 3} \frac{|x - 3|}{x - 3}$.

For $x \lt 3$: $\frac{|x-3|}{x-3} = \frac{3-x}{x-3} = -1$.

For $x \gt 3$: $\frac{|x-3|}{x-3} = \frac{x-3}{x-3} = 1$.

Left limit $= -1$Right limit $= 1$. The limit does not exist.

Proof that $\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1$.

This is proved using the squeeze theorem and the geometric inequality $\sin x \lt x \lt \tan x$ for $0 \lt x \lt \frac{\pi}{2}$Which gives $\cos x \lt \frac{\sin x}{x} \lt 1$.

As $x \to 0^+$, $\cos x \to 1$So by the squeeze theorem, $\frac{\sin x}{x} \to 1$. A similar Argument applies from the left. $\blacksquare$

Proof of the geometric inequality $\sin x \lt x \lt \tan x$ for $0 \lt x \lt \frac{\pi}{2}$. Consider a unit circle sector with angle $x$. The area of triangle $OAP$ (with altitude $\sin x$) is $\frac{1}{2}\sin x$The area of the sector is $\frac{1}{2}x$And the area of triangle $OAT$ (with Altitude $\tan x$) is $\frac{1}{2}\tan x$. Since the sector contains the first triangle and is Contained in the second, we get $\frac{1}{2}\sin x \lt \frac{1}{2}x \lt \frac{1}{2}\tan x$Hence $\sin x \lt x \lt \tan x$. $\blacksquare$

Proof that $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.

Let $h = e^x - 1$So $e^x = 1 + h$ and $x = \ln(1+h)$. As $x \to 0$, $h \to 0$.

Since $\displaystyle\lim_{h \to 0} \frac{\ln(1+h)}{h} = 1$ (which follows from $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ and the fact that $\ln$ and $\exp$ are Inverses), we obtain the result. $\blacksquare$

Proof that $\displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$.

Let $u = \ln(1+x)$So $e^u = 1 + x$ and $x = e^u - 1$. As $x \to 0$, $u \to 0$.

Since $\displaystyle\lim_{u \to 0} \frac{e^u - 1}{u} = 1$The result follows. $\blacksquare$

If $g(x) \le f(x) \le h(x)$ for all $x$ near $a$ (except possibly at $a$), and:

Then $\displaystyle\lim_{x \to a} f(x) = L$.

Intuition: If $f$ is sandwiched between two functions that both approach $L$Then $f$ must also Approach $L$. The squeeze theorem is particularly useful when $f$ oscillates or is otherwise hard to Evaluate directly.

Example

Show that $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$.

Since $-1 \le \sin\!\left(\frac{1}{x}\right) \le 1$We have $-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2$.

Both $\displaystyle\lim_{x \to 0}(-x^2) = 0$ and $\displaystyle\lim_{x \to 0} x^2 = 0$.

By the squeeze theorem, $\displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0$.

Example

Show that $\displaystyle\lim_{x \to 0} x\cos\!\left(\frac{1}{x}\right) = 0$.

Since $-1 \le \cos\!\left(\frac{1}{x}\right) \le 1$We have $-|x| \le x\cos\!\left(\frac{1}{x}\right) \le |x|$.

Both $\displaystyle\lim_{x \to 0}(-|x|) = 0$ and $\displaystyle\lim_{x \to 0}|x| = 0$.

By the squeeze theorem, the limit is $0$.

Example

Show that $\displaystyle\lim_{x \to 0} x^2 e^{\sin(1/x)} = 0$.

Since $-1 \le \sin(1/x) \le 1$We have $e^{-1} \le e^{\sin(1/x)} \le e^1$.

Therefore:

Both $\displaystyle\lim_{x \to 0} e^{-1} x^2 = 0$ and $\displaystyle\lim_{x \to 0} e \cdot x^2 = 0$.

By the squeeze theorem, the limit is $0$.

If $\displaystyle\lim_{x \to a} f(x) = L$ and $\displaystyle\lim_{x \to a} g(x) = M$Then:

1. $\displaystyle\lim_{x \to a} [f(x) + g(x)] = L + M$
2. $\displaystyle\lim_{x \to a} [f(x) - g(x)] = L - M$
3. $\displaystyle\lim_{x \to a} [c \cdot f(x)] = cL$
4. $\displaystyle\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$
5. $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$Provided $M \ne 0$

Theorem (Limit of a power). If $\displaystyle\lim_{x \to a} f(x) = L$ and $n$ is a positive Integer, then $\displaystyle\lim_{x \to a} [f(x)]^n = L^n$.

Theorem (Limit of a root). If $\displaystyle\lim_{x \to a} f(x) = L$ and $n$ is a positive Integer, and $L \ge 0$ when $n$ is even, then $\displaystyle\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$.

Proof of property 1 (sum rule). We need to show that for every $\epsilon \gt 0$There exists $\delta \gt 0$ such that $|x - a| \lt \delta$ implies $|(f+g)(x) - (L+M)| \lt \epsilon$.

Note that $|(f+g)(x) - (L+M)| = |(f(x) - L) + (g(x) - M)| \le |f(x) - L| + |g(x) - M|$.

Choose $\delta_1$ so that $|x-a| \lt \delta_1$ implies $|f(x)-L| \lt \epsilon/2$. Choose $\delta_2$ So that $|x-a| \lt \delta_2$ implies $|g(x)-M| \lt \epsilon/2$. Let $\delta = \min(\delta_1, \delta_2)$. Then $|x-a| \lt \delta$ implies both bounds hold, so $|(f+g)(x) - (L+M)| \lt \epsilon/2 + \epsilon/2 = \epsilon$. $\blacksquare$

When a function is continuous at $a$The limit equals the function value:

For rational functions $\displaystyle\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials:

• If $\deg P \lt \deg Q$: $\displaystyle\lim_{x \to \pm\infty} \frac{P(x)}{Q(x)} = 0$
• If $\deg P = \deg Q$: \displaystyle\lim_{x \to \pm\infty} \frac{P(x)}{Q(x)} = \frac{\mathrm{leading coeff of P}{\mathrm{leading coeff of Q}
• If $\deg P \gt \deg Q$: the limit is $\pm\infty$

Why this works. For large $x$The leading term dominates. Dividing numerator and denominator by The highest power of $x$ in the denominator, all lower-order terms vanish.

Example

Find $\displaystyle\lim_{x \to \infty} \frac{3x^2 - 5x + 2}{7x^2 + x - 1}$.

Since both polynomials are degree 2, the limit equals the ratio of leading coefficients:

Example

Find $\displaystyle\lim_{x \to \infty} \frac{5x^3 - 2x + 1}{4x^2 + 3x}$.

Since $\deg P = 3 \gt \deg Q = 2$The limit is $+\infty$.

Example

Find $\displaystyle\lim_{x \to -\infty} \frac{2x^3 + x^2 - 5}{5x^3 - 3x + 2}$.

Both polynomials are degree 3. The limit equals the ratio of leading coefficients:

This confirms that the same shortcut works for $x \to -\infty$ when the degrees are equal.

When direct substitution yields $\frac{0}{0}$Algebraic manipulation is required.

Example

Find $\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$.

Factor the numerator:

Example

Find $\displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x - 1}$.

Factor using difference of cubes: $x^3 - 1 = (x - 1)(x^2 + x + 1)$.

Example

Find $\displaystyle\lim_{x \to 1} \frac{x^4 - 1}{x^2 - 1}$.

For $x \ne \pm 1$. Therefore:

For expressions involving radicals, multiply by the conjugate.

Example

Find $\displaystyle\lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x}$.

Multiply numerator and denominator by the conjugate $\sqrt{x+4} + 2$:

Example

Find $\displaystyle\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{x}$.

Multiply by $\frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$:

Example

Find $\displaystyle\lim_{x \to 5} \frac{\sqrt{x+4} - 3}{x - 5}$.

Multiply by the conjugate $\sqrt{x+4} + 3$:

Example

Find $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Using the identity $1 - \cos x = 2\sin^2\frac{x}{2}$:

Example

Find $\displaystyle\lim_{x \to 0} \frac{\tan x}{x}$.

Example

Find $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{x}$.

Rewrite to use the standard limit:

When the limit involves a fraction within a fraction, combine the numerator into a single fraction First.

Example

Find $\displaystyle\lim_{x \to 2} \frac{\frac{1}{x} - \frac{1}{2}}{x - 2}$.

Combine the numerator:

Therefore:

The precise definition: $\displaystyle\lim_{x \to a} f(x) = L$ means that for every $\epsilon \gt 0$There exists a $\delta \gt 0$ such that:

Intuition. Think of it as a game. Your opponent picks $\epsilon$ (how close $f(x)$ must be to $L$). You must respond with $\delta$ (how close $x$ must be to $a$). If you can always win this Game, the limit exists.

Example

Prove that $\displaystyle\lim_{x \to 3} (2x - 1) = 5$.

We need to show that for every $\epsilon \gt 0$There exists $\delta \gt 0$ such that $0 \lt |x - 3| \lt \delta \implies |(2x-1) - 5| \lt \epsilon$.

Working backwards: $|(2x-1) - 5| = |2x - 6| = 2|x - 3|$.

We want $2|x - 3| \lt \epsilon$So choose $\delta = \frac{\epsilon}{2}$.

Proof. Let $\epsilon \gt 0$. Choose $\delta = \frac{\epsilon}{2}$. Then:

Therefore, $\displaystyle\lim_{x \to 3} (2x - 1) = 5$. $\blacksquare$

Example

Prove that $\displaystyle\lim_{x \to 2} x^2 = 4$.

We need $|x^2 - 4| \lt \epsilon$ whenever $0 \lt |x - 2| \lt \delta$.

Note that $|x^2 - 4| = |x - 2| \cdot |x + 2|$.

If we restrict $\delta \le 1$Then $|x - 2| \lt 1$So $1 \lt x \lt 3$ and $|x + 2| \lt 5$.

Thus $|x^2 - 4| = |x - 2| \cdot |x + 2| \lt 5|x - 2|$.

Choose $\delta = \min\!\left(1, \frac{\epsilon}{5}\right)$.

Proof. Let $\epsilon \gt 0$. Choose $\delta = \min\!\left(1, \frac{\epsilon}{5}\right)$. If $0 \lt |x - 2| \lt \delta$Then:

Therefore, $\displaystyle\lim_{x \to 2} x^2 = 4$. $\blacksquare$

Example

Prove that $\displaystyle\lim_{x \to a} \sqrt{x} = \sqrt{a}$ for $a \gt 0$.

We need $|\sqrt{x} - \sqrt{a}| \lt \epsilon$ whenever $0 \lt |x - a| \lt \delta$.

Rationalise: $|\sqrt{x} - \sqrt{a}| = \frac{|x - a|}{\sqrt{x} + \sqrt{a}} \le \frac{|x - a|}{\sqrt{a}}$.

We want $\frac{|x - a|}{\sqrt{a}} \lt \epsilon$So $|x - a| \lt \epsilon\sqrt{a}$.

Choose $\delta = \min(a, \epsilon\sqrt{a})$. The condition $\delta \le a$ ensures $x \gt 0$ so that $\sqrt{x}$ is defined.

Proof. Let $\epsilon \gt 0$. Choose $\delta = \min(a, \epsilon\sqrt{a})$. If $0 \lt |x - a| \lt \delta$Then $x \gt 0$ and:

Therefore, $\displaystyle\lim_{x \to a} \sqrt{x} = \sqrt{a}$. $\blacksquare$

The general approach is:

1. Start with $|f(x) - L|$ and try to bound it in terms of $|x - a|$.
2. If $f$ involves products, use the “restrict delta” technique: bound each factor separately.
3. If $f$ involves roots, rationalise and use the fact that $\sqrt{x} + \sqrt{a} \ge \sqrt{a}$.
4. Choose \delta = \min(\mathrm{bound, \epsilon / \mathrm{constant) to handle both the restriction and the $\epsilon$ requirement.

A function $f$ is continuous at $a$ if all three conditions hold:

1. $f(a)$ is defined
2. $\displaystyle\lim_{x \to a} f(x)$ exists
3. $\displaystyle\lim_{x \to a} f(x) = f(a)$

Theorem. Every polynomial function is continuous everywhere. Every rational function is Continuous on its domain.

Theorem (Continuity of compositions). If $g$ is continuous at $a$ and $f$ is continuous at $g(a)$Then $f \circ g$ is continuous at $a$.

This theorem justifies statements like ”$\sqrt{x^2 + 1}$ is continuous everywhere” — $x^2 + 1$ is a Polynomial (continuous everywhere) and $\sqrt{x}$ is continuous at all positive values (and $x^2 + 1 \ge 1 \gt 0$).

If $f$ is continuous on $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$Then there exists At least one $c \in (a, b)$ such that $f(c) = k$.

Example

Show that $f(x) = x^3 + x - 1$ has a root in $(0, 1)$.

$f(0) = -1 \lt 0$ and $f(1) = 1 \gt 0$.

Since $f$ is continuous on $[0, 1]$ and $0$ is between $f(0)$ and $f(1)$By the IVT there exists $c \in (0, 1)$ such that $f(c) = 0$.

Application of IVT to bisection. The IVT motivates the bisection method for root-finding: if $f(a)$ and $f(b)$ have opposite signs, a root exists in $(a, b)$. Halving the interval and checking Signs converges to the root.

Example

Show that $f(x) = e^x - 3 - x$ has at least one root in $(1, 2)$.

$f(1) = e - 4 \approx -1.282 \lt 0$ and $f(2) = e^2 - 5 \approx 2.389 \gt 0$.

Since $f$ is continuous (as a sum of continuous functions) on $[1, 2]$By the IVT there exists $c \in (1, 2)$ such that $f(c) = 0$. $\blacksquare$

Corollary of the IVT. If $f$ is continuous on $[a, b]$ and $f(a) \cdot f(b) \lt 0$Then $f$ has At least one zero in $(a, b)$.

Example

Find the asymptotes of $\displaystyle f(x) = \frac{2x^2 + 3x - 1}{x + 1}$.

Vertical asymptote: Set denominator to zero: $x + 1 = 0 \implies x = -1$.

Slant asymptote: Perform long division:

The slant asymptote is $y = 2x + 1$.

Example

Find the horizontal asymptotes of $\displaystyle f(x) = \frac{3e^x}{e^x + 1}$.

As $x \to \infty$: Divide numerator and denominator by $e^x$:

As $x \to -\infty$: Divide numerator and denominator by $e^x$:

Horizontal asymptotes: $y = 3$ (as $x \to \infty$) and $y = 0$ (as $x \to -\infty$).

If $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)}$ produces the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$And $f$ and $g$ are differentiable near $a$ with $g'(x) \ne 0$ near $a$Then:

Provided the limit on the right exists.

When to use L’Hopital’s Rule. It applies ONLY to $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ Forms. Using it on a determinate form (e.g., $\frac{3}{5}$) is an error.

When L’Hopital’s Rule fails. If the limit $\displaystyle\lim_{x \to a} \frac{f'(x)}{g'(x)}$ does Not exist, this does NOT mean the original limit does not exist. L’Hopital’s Rule only gives a Conclusion when the right-hand limit exists (or is $\pm\infty$).

Example

Find $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x}$.

Direct substitution gives $\frac{0}{0}$. Apply L’Hopital’s Rule:

Example

Find $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Direct substitution gives $\frac{0}{0}$:

Note that we applied L’Hopital’s Rule twice, since the second attempt still gave $\frac{0}{0}$.

Example

Find $\displaystyle\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}$.

This is $\frac{\infty}{\infty}$. Apply L’Hopital’s Rule:

Example

Find $\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3}$.

Direct substitution gives $\frac{0}{0}$. Apply L’Hopital’s Rule three times:

Example

Find $\displaystyle\lim_{x \to 0^+} x \ln x$.

This has the form $0 \cdot (-\infty)$Which is indeterminate. Rewrite as a quotient:

Now it is $\frac{-\infty}{\infty}$. Apply L’Hopital’s Rule:

Example

Let $f(x) = \begin{cases} x^2 + 1 & x \lt 2 \\ 3x - 1 & x \ge 2 \end{cases}$.

Find $\displaystyle\lim_{x \to 2} f(x)$ and determine if $f$ is continuous at $x = 2$.

Left-hand limit: $\displaystyle\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 + 1) = 5$.

Right-hand limit: $\displaystyle\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3x - 1) = 5$.

Since both one-sided limits equal 5: $\displaystyle\lim_{x \to 2} f(x) = 5$.

Check continuity: $f(2) = 3(2) - 1 = 5 = \lim_{x \to 2} f(x)$.

Therefore, $f$ is continuous at $x = 2$.

Example

Let $g(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x \neq 2 \\ k & x = 2 \end{cases}$.

Find $k$ such that $g$ is continuous at $x = 2$.

$\displaystyle\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = 4$.

For continuity: $k = g(2) = 4$.

Example

Let $h(x) = \begin{cases} x^2 + bx + 1 & x \le 0 \\ 2x + 3 & x \gt 0 \end{cases}$.

Find $b$ such that $h$ is continuous at $x = 0$.

Left-hand limit: $\displaystyle\lim_{x \to 0^-} h(x) = 0 + 0 + 1 = 1$.

Right-hand limit: $\displaystyle\lim_{x \to 0^+} h(x) = 3$.

For continuity: $1 = 3$Which is impossible. No value of $b$ makes $h$ continuous at $x = 0$.

This example demonstrates that continuity at a junction point of a piecewise function is not always Achievable — on whether the one-sided limits can be made to agree.