Some limits arise so often that they are worth committing to memory as standard results. Each is itself a $\dfrac{0}{0}$ form that cannot be cracked by simple cancellation, yet each settles to a clean value. Knowing them turns long derivations into one-line substitutions.
The trigonometric limit. The cornerstone of calculus with angles — valid only when $x$ is measured in radians:
$$\lim_{x \to 0}\dfrac{\sin x}{x} = 1$$
An immediate companion follows from it: $\lim_{x \to 0}\dfrac{\tan x}{x} = 1$, since $\tan x = \dfrac{\sin x}{\cos x}$ and $\cos x \to 1$. A frequent supporting result is $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$, which follows from the identity $1 - \cos x = 2\sin^2\dfrac{x}{2}$; note that the closely related $\lim_{x \to 0}\dfrac{1 - \cos x}{x} = 0$ (one power of $x$ less in the denominator).
The exponential and logarithmic limits. These power the derivatives of $e^x$, $a^x$ and $\ln x$:
$$\lim_{x \to 0}\dfrac{e^x - 1}{x} = 1, \qquad \lim_{x \to 0}\dfrac{a^x - 1}{x} = \ln a \;\; (a > 0), \qquad \lim_{x \to 0}\dfrac{\log(1 + x)}{x} = 1$$
(The logarithm here is the natural log, base $e$.)
The power limit. A purely algebraic result that generalises difference-of-powers factorisation to any real index $n$:
$$\lim_{x \to a}\dfrac{x^n - a^n}{x - a} = n\,a^{\,n - 1}$$
Collected together for quick reference:
How to use them. The trick is to massage the expression until it matches a standard shape exactly. The key fact that licenses every such rewrite is: if $g(x) \to 0$ as $x \to 0$, then $\dfrac{\sin g(x)}{g(x)} \to 1$ as well. So to evaluate $\lim_{x \to 0}\dfrac{\sin 5x}{x}$, write it as $5 \cdot \dfrac{\sin 5x}{5x}$; as $x \to 0$ the angle $5x \to 0$ too, so $\dfrac{\sin 5x}{5x} \to 1$ and the limit is $5$. For a ratio of two sines, force each over its own angle and let the leftover coefficients give the answer — thus $\lim_{x \to 0}\dfrac{\sin ax}{\sin bx} = \dfrac{a}{b}$.
Deeper Insight — why radians, and why these limits are the seeds of differentiation: The result $\lim_{x \to 0}\tfrac{\sin x}{x}=1$ holds only in radians, and that is no accident: the radian is defined precisely so that, for tiny angles, the arc length and the vertical sine are practically equal, which is what makes the ratio tend to $1$. Switch to degrees and you would instead get $\tfrac{\pi}{180}$, contaminating every trigonometric derivative — this is the deep reason calculus always uses radians. More broadly, every standard limit here is really a derivative in disguise evaluated at a single point: $\lim_{x\to 0}\tfrac{e^x-1}{x}$ is the slope of $e^x$ at $x=0$, and $\lim_{x\to a}\tfrac{x^n-a^n}{x-a}$ is the slope of $x^n$ at $x=a$. Recognising them as building blocks, rather than isolated facts to memorise, is what lets you derive the rules of the next topic instead of merely recalling them.
Evaluate $\lim_{x \to 0}\dfrac{\sin 5x}{x}$.
Solution- Reshape to match $\dfrac{\sin(\cdot)}{(\cdot)}$: multiply and divide by $5$.
- $\dfrac{\sin 5x}{x} = 5 \cdot \dfrac{\sin 5x}{5x}$.
- As $x \to 0$, the angle $5x \to 0$, so $\dfrac{\sin 5x}{5x} \to 1$.
- Therefore the limit is $5 \times 1 = 5$.
Answer: $\lim_{x \to 0}\dfrac{\sin 5x}{x} = 5$.
Evaluate $\lim_{x \to 0}\dfrac{\sin 3x}{\sin 7x}$.
Solution- Force each sine over its own angle by inserting $3x$ and $7x$.
- $\dfrac{\sin 3x}{\sin 7x} = \dfrac{\sin 3x}{3x} \cdot \dfrac{7x}{\sin 7x} \cdot \dfrac{3x}{7x}$.
- As $x \to 0$: $\dfrac{\sin 3x}{3x} \to 1$ and $\dfrac{7x}{\sin 7x} \to 1$, leaving $\dfrac{3}{7}$.
Answer: $\lim_{x \to 0}\dfrac{\sin 3x}{\sin 7x} = \dfrac{3}{7}$.
Evaluate $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2}$.
Solution- Use the identity $1 - \cos x = 2\sin^2\dfrac{x}{2}$.
- $\dfrac{2\sin^2(x/2)}{x^2} = \dfrac{2\sin^2(x/2)}{4\,(x/2)^2} = \dfrac{1}{2}\left(\dfrac{\sin(x/2)}{x/2}\right)^2$.
- As $x \to 0$, $\dfrac{\sin(x/2)}{x/2} \to 1$, so the bracket squared $\to 1$.
- Limit $= \dfrac{1}{2} \times 1 = \dfrac{1}{2}$.
Answer: $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$.
Evaluate $\lim_{x \to 0}\dfrac{e^{2x} - 1}{x}$.
Solution- Reshape to match $\dfrac{e^{(\cdot)} - 1}{(\cdot)}$ by inserting $2x$.
- $\dfrac{e^{2x} - 1}{x} = 2 \cdot \dfrac{e^{2x} - 1}{2x}$.
- As $x \to 0$, $2x \to 0$, so $\dfrac{e^{2x} - 1}{2x} \to 1$.
- Limit $= 2 \times 1 = 2$.
Answer: $\lim_{x \to 0}\dfrac{e^{2x} - 1}{x} = 2$.
Evaluate $\lim_{x \to 2}\dfrac{x^5 - 32}{x - 2}$.
Solution- This is the standard form $\lim_{x \to a}\dfrac{x^n - a^n}{x - a} = n\,a^{\,n-1}$ with $a = 2$ (since $32 = 2^5$).
- Here $n = 5$ and $a = 2$.
- Apply the formula: $n\,a^{\,n-1} = 5 \cdot 2^{4} = 5 \times 16 = 80$.
Answer: $\lim_{x \to 2}\dfrac{x^5 - 32}{x - 2} = 80$.
Evaluate $\lim_{x \to 0}\dfrac{3^x - 1}{x}$.
Solution- This is the standard exponential form $\lim_{x \to 0}\dfrac{a^x - 1}{x} = \ln a$.
- Here $a = 3$.
- So the limit equals $\ln 3$.
Answer: $\lim_{x \to 0}\dfrac{3^x - 1}{x} = \ln 3$.
Evaluate $\lim_{x \to 0}\dfrac{\tan 4x}{x}$.
Solution- Write $\dfrac{\tan 4x}{x} = 4 \cdot \dfrac{\tan 4x}{4x}$.
- As $x \to 0$, the angle $4x \to 0$, so $\dfrac{\tan 4x}{4x} \to 1$.
- Limit $= 4 \times 1 = 4$.
Answer: $\lim_{x \to 0}\dfrac{\tan 4x}{x} = 4$.
Evaluate $\lim_{x \to 0}\dfrac{1 - \cos 2x}{x^2}$.
Solution- Use $1 - \cos 2x = 2\sin^2 x$.
- $\dfrac{2\sin^2 x}{x^2} = 2\left(\dfrac{\sin x}{x}\right)^2$.
- As $x \to 0$, $\dfrac{\sin x}{x} \to 1$, so the square $\to 1$.
- Limit $= 2 \times 1 = 2$.
Answer: $\lim_{x \to 0}\dfrac{1 - \cos 2x}{x^2} = 2$.
Evaluate $\lim_{x \to 0}\dfrac{\log(1 + 3x)}{x}$.
Solution- Match the standard form $\dfrac{\log(1 + (\cdot))}{(\cdot)}$ by inserting $3x$.
- $\dfrac{\log(1 + 3x)}{x} = 3 \cdot \dfrac{\log(1 + 3x)}{3x}$.
- As $x \to 0$, $3x \to 0$, so $\dfrac{\log(1 + 3x)}{3x} \to 1$.
- Limit $= 3 \times 1 = 3$.
Answer: $\lim_{x \to 0}\dfrac{\log(1 + 3x)}{x} = 3$.
Evaluate $\lim_{x \to 3}\dfrac{x^4 - 81}{x - 3}$.
Solution- Recognise $81 = 3^4$, so this is $\lim_{x \to a}\dfrac{x^n - a^n}{x - a}$ with $n = 4$, $a = 3$.
- Apply $n\,a^{\,n-1}$.
- $= 4 \cdot 3^{3} = 4 \times 27 = 108$.
Answer: $\lim_{x \to 3}\dfrac{x^4 - 81}{x - 3} = 108$.
Evaluate $\lim_{x \to 0}\dfrac{\sin 2x + \sin 6x}{4x}$.
Solution- Split the fraction: $\dfrac{\sin 2x}{4x} + \dfrac{\sin 6x}{4x}$.
- Reshape each: $\dfrac{2}{4}\cdot\dfrac{\sin 2x}{2x} + \dfrac{6}{4}\cdot\dfrac{\sin 6x}{6x}$.
- As $x \to 0$ both reshaped ratios $\to 1$, leaving $\dfrac{2}{4} + \dfrac{6}{4} = \dfrac{8}{4} = 2$.
Answer: $\lim_{x \to 0}\dfrac{\sin 2x + \sin 6x}{4x} = 2$.