The standard limits

What you will be able to do

• state the standard limits of sine, tangent, cosine, powers, exponentials and logarithms;
• bend a given limit into a standard one by multiplying and dividing;
• see where each standard limit comes from.

Why we need them

Factorising cannot help with $\dfrac{\sin x}{x}$ or $\dfrac{e^x - 1}{x}$. These need new facts. We state them, explain where they come from, and then use them again and again. In every one, $x$ is a pure number, which for angles means radians.

The sine limit

The idea is a picture. On a circle of radius $1$, an angle $x$ (in radians, with $0 < x < \dfrac{\pi}{2}$) gives an arc of length $x$. The arc is longer than the vertical side $\sin x$ and shorter than the tangent length $\tan x$. So

Divide by $\sin x$ and take reciprocals. This gives

As $x \to 0$ the cosine tends to $1$, so $\dfrac{\sin x}{x}$ is squeezed towards $1$. Course 3 names this the squeeze theorem.

Three relatives follow at once:

• $\displaystyle\lim_{x \to 0} \dfrac{\tan x}{x} = 1$, since $\dfrac{\tan x}{x} = \dfrac{\sin x}{x} \cdot \dfrac{1}{\cos x}$;
• $\displaystyle\lim_{x \to 0} \dfrac{\sin kx}{x} = k$, since $\dfrac{\sin kx}{x} = k \cdot \dfrac{\sin kx}{kx}$;
• $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$, since $1 - \cos x = 2\sin^2\dfrac{x}{2}$.

Degrees. If $x$ is in degrees, then $\dfrac{\sin x}{x} \to \dfrac{\pi}{180}$ instead. All of calculus uses radians, in India, the UK, the US and everywhere else, so we always do.

The power limit

For a whole number $n$ the proof is the factorisation $x^n - a^n = (x - a)(x^{n-1} + x^{n-2}a + \dots + a^{n-1})$. After cancelling $x - a$, the second bracket has $n$ terms, and each tends to $a^{n-1}$. The result also holds for a fraction $n$, such as $n = \dfrac{1}{2}$, when $a > 0$.

The exponential and the logarithm

The number $e \approx 2.718$ is chosen so that the curve $y = e^x$ leaves the point $(0, 1)$ with slope exactly $1$. In limit language:

The third follows from the first. Write $a^x = e^{x \ln a}$, and put $t = x\ln a$. Then $\dfrac{a^x - 1}{x} = \ln a \cdot \dfrac{e^t - 1}{t} \to \ln a$. Here $\ln$ means the logarithm to base $e$, which some books write as $\log_e$. In many Indian textbooks a plain $\log$ also means $\ln$ in calculus.

A worked example

Find the following limits.

1. $\displaystyle\lim_{x \to 0} \dfrac{\sin 3x}{\sin 5x}$. Divide the top and the bottom by $x$: $\dfrac{\sin 3x / x}{\sin 5x / x} \to \dfrac{3}{5}$.
2. $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos 4x}{x^2}$. Write $\dfrac{1 - \cos 4x}{x^2} = 16 \cdot \dfrac{1 - \cos 4x}{(4x)^2}$. The last fraction tends to $\dfrac{1}{2}$, so the limit is $8$.
3. $\displaystyle\lim_{x \to 2} \dfrac{x^5 - 32}{x - 2}$. Here $32 = 2^5$, so the limit is $5 \times 2^4 = 80$.
4. $\displaystyle\lim_{x \to 0} \dfrac{e^{3x} - 1}{x} = 3 \cdot \dfrac{e^{3x} - 1}{3x} \to 3$, and $\displaystyle\lim_{x \to 0} \dfrac{\ln(1 + 5x)}{x} = 5$.

The skill is the same each time: multiply and divide so that the inside of the function matches the denominator.

Practise with new numbers

Try it in Python

The program prints six ratios at $x = 0.1,\ 0.01,\ 0.001$. Each one moves towards its standard value.

Common mistakes

• Using degrees. The standard limit of $\dfrac{\sin x}{x}$ is $1$ for $x$ in radians only.
• Mismatching the inside. $\dfrac{\sin 3x}{x}$ is not $1$. Make the inside and the bottom equal: $3 \cdot \dfrac{\sin 3x}{3x}$.
• Dropping the power. $\dfrac{x^n - a^n}{x - a}$ gives $na^{n-1}$, with the power reduced by $1$.
• Mixing $\ln a$ with $a$. $\dfrac{a^x - 1}{x}$ tends to $\ln a$, not to $a$.

Remember these

• $\dfrac{\sin x}{x} \to 1$, $\dfrac{\tan x}{x} \to 1$, $\dfrac{1 - \cos x}{x^2} \to \dfrac{1}{2}$ as $x \to 0$, with $x$ in radians.
• $\dfrac{x^n - a^n}{x - a} \to na^{n-1}$ as $x \to a$.
• $\dfrac{e^x - 1}{x} \to 1$, $\dfrac{\ln(1 + x)}{x} \to 1$, $\dfrac{a^x - 1}{x} \to \ln a$ as $x \to 0$.
• Multiply and divide to make the inside match the denominator.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\lim_{x \to 0} \dfrac{\sin 7x}{x}$.

2. Find $\displaystyle\lim_{x \to 0} \dfrac{\tan 2x}{3x}$.

3. Find $\displaystyle\lim_{x \to 3} \dfrac{x^4 - 81}{x - 3}$.

4. Find $\displaystyle\lim_{x \to 0} \dfrac{5^x - 1}{x}$.

5. A student says that for $x$ in degrees, $\dfrac{\sin x}{x} \to 1$. What is wrong?

6. Find $\displaystyle\lim_{x \to 0} \dfrac{\ln(1 + 4x)}{2x}$.

References

• OpenStax, Calculus Volume 1, Chapter 2, section 2.3 on the limits of trigonometric functions and section 3.5 for the derivative of sine.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, the section on limits of trigonometric functions. The exponential and logarithmic limits belong to the JEE Main syllabus and to A-Level and IB texts.