Fundamental Limits

After learning the properties of limits and some calculation techniques, we can study some limits that frequently appear in Calculus I.

These limits are called fundamental limits because they serve as the basis for solving many other limits and, later, for developing derivatives.

The main results we will study are:

$$\lim_{x\to0}\frac{\operatorname{sin}x}{x}=1$$

and

$$\lim_{x\to0}\frac{1-\cos x}{x}=0.$$

We will also see how to use these results to calculate other trigonometric limits.

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1. The Fundamental Sine Limit

One of the most important limits in Calculus is

$$\boxed{ \lim_{x\to0}\frac{\operatorname{sin}x}{x}=1 }$$

This result is valid when $x$ is measured in radians.

This observation is fundamental.

The limit $\displaystyle\lim_{x\to0}\frac{\operatorname{sin}x}{x}=1$ has this form only when the angle is expressed in radians.

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2. An Intuitive Interpretation

As $x$ approaches $0$, we also have

$$\operatorname{sin}x\to0.$$

Therefore, initially we have an expression of the form

$$\frac00.$$

However, even though the numerator and denominator both approach zero, their ratio approaches $1$:

$$\frac{\operatorname{sin}x}{x}\to1.$$

We can observe this numerically.

$x$	$\frac{\operatorname{sin}x}{x}$
$0{,}5$	approximately $0{,}9589$
$0{,}1$	approximately $0{,}9983$
$0{,}01$	approximately $0{,}99998$
$0{,}001$	approximately $0{,}9999998$

As $x$ approaches $0$, the ratio approaches $1$.

Thus,

$$\boxed{ \lim_{x\to0}\frac{\operatorname{sin}x}{x}=1 }$$

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The Fundamental Inequality

Consider $\theta$ such that

$$0<\theta<\frac{\pi}{2}.$$

From the geometric interpretation of the sine and tangent functions on the unit circle, we have

By comparing the areas of the three geometric elements, we have

$$\frac{\cos\theta\operatorname{sin}\theta}{2} < \frac{\theta}{2} < \frac{\tan\theta}{2}.$$

Since

$$0<\theta<\frac{\pi}{2},$$

we have

$$\operatorname{sin}\theta>0 \qquad\text{and}\qquad \cos\theta>0.$$

Dividing the entire inequality by

$$\frac{\operatorname{sin}\theta}{2},$$

we obtain

$$\cos\theta < \frac{\theta}{\operatorname{sin}\theta} < \frac{1}{\cos\theta}.$$

Since all quantities are positive, we can invert the inequalities. Thus,

$$\frac{1}{\cos\theta} > \frac{\operatorname{sin}\theta}{\theta} > \cos\theta.$$

Rearranging the terms,

$$\cos\theta < \frac{\operatorname{sin}\theta}{\theta} < \frac{1}{\cos\theta}.$$

On the other hand, from the initial inequality

$$\operatorname{sin}\theta<\theta,$$

and since $\theta>0$, we have

$$\frac{\operatorname{sin}\theta}{\theta}<1.$$

Therefore,

$$\boxed{ \cos\theta < \frac{\operatorname{sin}\theta}{\theta} < 1 }.$$

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Analyzing Negative Values

We also need to analyze what happens when $\theta$ approaches $0$ from the left.

Consider

$$-\frac{\pi}{2}<\theta<0.$$

Then,

$$0<-\theta<\frac{\pi}{2}.$$

We can apply the inequality obtained previously to the number $-\theta$:

$$\cos(-\theta) < \frac{\operatorname{sin}(-\theta)}{-\theta} < 1.$$

We know that cosine is an even function:

$$\cos(-\theta)=\cos\theta,$$

and sine is an odd function:

$$\operatorname{sin}(-\theta)=-\operatorname{sin}\theta.$$

Therefore,

$$\frac{\operatorname{sin}(-\theta)}{-\theta} = \frac{-\operatorname{sin}\theta}{-\theta} = \frac{\operatorname{sin}\theta}{\theta}.$$

Thus,

$$\cos\theta < \frac{\operatorname{sin}\theta}{\theta} < 1.$$

Therefore, the same inequality holds when

$$-\frac{\pi}{2}<\theta<0.$$

Consequently, for every $\theta$ such that

$$0<|\theta|<\frac{\pi}{2},$$

we have

$$\boxed{ \cos\theta < \frac{\operatorname{sin}\theta}{\theta} < 1 }.$$

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Letting $\theta$ Approach Zero

Now we can calculate the limit

$$\lim_{\theta\to0} \frac{\operatorname{sin}\theta}{\theta}.$$

We know that, for $\theta$ sufficiently close to $0$, with $\theta\neq0$,

$$\cos\theta < \frac{\operatorname{sin}\theta}{\theta} < 1.$$

Let's calculate the limits of the expressions that appear at the endpoints.

Since the cosine function is continuous,

$$\lim_{\theta\to0}\cos\theta = \cos0 = 1.$$

Furthermore,

$$\lim_{\theta\to0}1=1.$$

Therefore,

$$\lim_{\theta\to0}\cos\theta = 1 = \lim_{\theta\to0}1.$$

Thus, we have

$$\cos\theta < \frac{\operatorname{sin}\theta}{\theta} < 1,$$

with both endpoints approaching $1$ as $\theta\to0$.

By the Squeeze Theorem, we conclude that

$$\boxed{ \lim_{\theta\to0} \frac{\operatorname{sin}\theta}{\theta} = 1 }.$$

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Conclusion

We have geometrically demonstrated that

$$\boxed{ \lim_{x\to0} \frac{\operatorname{sin}x}{x} = 1 }.$$

This result is known as the fundamental trigonometric limit.

It is important to note that the equality depends on using radians to measure angles.

From this limit, we can obtain several other important results, such as

$$\lim_{x\to0}\frac{\operatorname{sin}(ax)}{x}=a,$$

and

$$\lim_{x\to0}\frac{\tan x}{x}=1.$$

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3. Consequence: $\displaystyle\lim_{x\to0}\frac{x}{\operatorname{sin}x}$

We know that

$$\lim_{x\to0}\frac{\operatorname{sin}x}{x}=1.$$

Taking the reciprocal:

$$\lim_{x\to0}\frac{x}{\operatorname{sin}x} = \frac{1}{1}.$$

Therefore,

$$\boxed{ \lim_{x\to0}\frac{x}{\operatorname{sin}x}=1 }$$

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4. Example

Calculate

$$\lim_{x\to0}\frac{\operatorname{sin}(5x)}{x}.$$

We want to make the fundamental limit appear.

We multiply and divide by $5$:

$$\frac{\operatorname{sin}(5x)}{x} = 5\frac{\operatorname{sin}(5x)}{5x}.$$

Therefore,

$$\lim_{x\to0}\frac{\operatorname{sin}(5x)}{x} = 5\lim_{x\to0} \frac{\operatorname{sin}(5x)}{5x}.$$

Let

$$u=5x,$$

when $x\to0$, we have $u\to0$.

Thus,

$$\lim_{x\to0} \frac{\operatorname{sin}(5x)}{5x} = 1.$$

Therefore,

$$\boxed{ \lim_{x\to0}\frac{\operatorname{sin}(5x)}{x}=5 }$$

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5. General Case

Similarly, for a constant $a$,

$$\boxed{ \lim_{x\to0} \frac{\operatorname{sin}(ax)}{x} = a }$$

provided that $a$ is constant.

Proof

We write

$$\frac{\operatorname{sin}(ax)}{x} = a\frac{\operatorname{sin}(ax)}{ax}.$$

Therefore,

$$\lim_{x\to0} \frac{\operatorname{sin}(ax)}{x} = a \lim_{x\to0} \frac{\operatorname{sin}(ax)}{ax}.$$

By the fundamental limit,

$$\lim_{x\to0} \frac{\operatorname{sin}(ax)}{ax} =1.$$

Therefore,

$$\boxed{ \lim_{x\to0} \frac{\operatorname{sin}(ax)}{x}=a }$$

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6. Fundamental Cosine Limit

Another important result is

$$\boxed{ \lim_{x\to0}\frac{1-\cos x}{x}=0 }$$

We can understand this result using a trigonometric identity.

We know that

$$1-\cos x = 2\operatorname{sin}^2\left(\frac{x}{2}\right).$$

Then,

$$\frac{1-\cos x}{x} = \frac{ 2\operatorname{sin}^2\left(\frac{x}{2}\right) }{x}.$$

We can write

$$x=2\frac{x}{2}.$$

Thus,

$$\frac{1-\cos x}{x} = \frac{ \operatorname{sin}^2\left(\frac{x}{2}\right) }{ \frac{x}{2} }.$$

Now we write:

$$\frac{ \operatorname{sin}^2\left(\frac{x}{2}\right) }{ \frac{x}{2} } = \operatorname{sin}\left(\frac{x}{2}\right) \frac{ \operatorname{sin}\left(\frac{x}{2}\right) }{ \frac{x}{2} }.$$

When $x\to0$,

$$\operatorname{sin}\left(\frac{x}{2}\right)\to0$$

and

$$\frac{ \operatorname{sin}\left(\frac{x}{2}\right) }{ \frac{x}{2} } \to1.$$

Therefore,

$$\boxed{ \lim_{x\to0}\frac{1-\cos x}{x}=0 }$$

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7. A Very Useful Result

We can obtain another important limit:

$$\lim_{x\to0}\frac{1-\cos x}{x^2}.$$

Using

$$1-\cos x = 2\operatorname{sin}^2\left(\frac{x}{2}\right),$$

we have

$$\frac{1-\cos x}{x^2} = \frac{ 2\operatorname{sin}^2\left(\frac{x}{2}\right) }{x^2}.$$

Since

$$x^2=4\left(\frac{x}{2}\right)^2,$$

we obtain

$$\frac{1-\cos x}{x^2} = \frac12 \left[ \frac{ \operatorname{sin}\left(\frac{x}{2}\right) }{ \frac{x}{2} } \right]^2.$$

Applying the fundamental limit:

$$\lim_{x\to0} \frac{ \operatorname{sin}\left(\frac{x}{2}\right) }{ \frac{x}{2} } =1.$$

Therefore,

$$\boxed{ \lim_{x\to0}\frac{1-\cos x}{x^2} = \frac12 }$$

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8. Tangent Limit

We can use

$$\tan x= \frac{\operatorname{sin}x}{\cos x}.$$

Thus,

$$\frac{\tan x}{x} = \frac{\operatorname{sin}x}{x\cos x}.$$

Separating the factors:

$$\frac{\tan x}{x} = \frac{\operatorname{sin}x}{x} \cdot \frac1{\cos x}.$$

When $x\to0$,

$$\frac{\operatorname{sin}x}{x}\to1$$

and

$$\cos x\to1.$$

Therefore,

$$\boxed{ \lim_{x\to0}\frac{\tan x}{x}=1 }$$

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9. Example with the Tangent

Calculate

$$\lim_{x\to0}\frac{\tan(3x)}{x}.$$

We write

$$\frac{\tan(3x)}{x} = 3\frac{\tan(3x)}{3x}.$$

By the fundamental tangent limit,

$$\lim_{x\to0} \frac{\tan(3x)}{3x} =1.$$

Therefore,

$$\boxed{ \lim_{x\to0}\frac{\tan(3x)}{x}=3 }$$

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10. A Limit Involving Sine and Cosine

Consider

$$\lim_{x\to0} \frac{\operatorname{sin}(4x)}{2x}.$$

We can write

$$\frac{\operatorname{sin}(4x)}{2x} = 2 \frac{\operatorname{sin}(4x)}{4x}.$$

Thus,

$$\lim_{x\to0} \frac{\operatorname{sin}(4x)}{2x} = 2.$$

Therefore,

$$\boxed{ \lim_{x\to0} \frac{\operatorname{sin}(4x)}{2x}=2 }$$

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11. Attention to Indeterminate Forms

It is important not to confuse an indeterminate form with the value of the limit.

For example,

$$\lim_{x\to0} \frac{\operatorname{sin}x}{x}$$

produces, by direct substitution,

$$\frac00.$$

However,

$$\boxed{ \lim_{x\to0} \frac{\operatorname{sin}x}{x}=1 }$$

Therefore,

$$\frac00$$

does not mean that the limit is zero.

It simply means that direct substitution is not sufficient.

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12. Summary of the Fundamental Limits

The most important results from this topic are:

$$\boxed{ \lim_{x\to0}\frac{\operatorname{sin}x}{x}=1 }$$ $$\boxed{ \lim_{x\to0}\frac{x}{\operatorname{sin}x}=1 }$$ $$\boxed{ \lim_{x\to0}\frac{1-\cos x}{x}=0 }$$ $$\boxed{ \lim_{x\to0}\frac{1-\cos x}{x^2} = \frac12 }$$ $$\boxed{ \lim_{x\to0}\frac{\tan x}{x}=1 }$$

And, for a constant $a$,

$$\boxed{ \lim_{x\to0}\frac{\operatorname{sin}(ax)}{x}=a }$$

and

$$\boxed{ \lim_{x\to0}\frac{\tan(ax)}{x}=a. }$$

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13. Strategy for Solving Trigonometric Limits

When encountering a limit involving trigonometric functions, try to transform the expression until one of the fundamental forms appears:

$$\frac{\operatorname{sin}u}{u}$$

or

$$\frac{\tan u}{u}.$$

For example:

$$\frac{\operatorname{sin}(5x)}{x} = 5\frac{\operatorname{sin}(5x)}{5x}.$$

From there, we can directly apply

$$\lim_{u\to0}\frac{\operatorname{sin}u}{u}=1.$$

This strategy will be frequently used in the study of derivatives.

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14. Next Step

The fundamental limits allow us to move on to one of the central concepts of Differential Calculus:

The continuity of a function.

In the next topic, we will study what it means for a function to be continuous at a point and how to use limits to verify continuity.