Limits of Trigonometric Functions

A limit of a trigonometric function describes the value that the function f(x) approaches as the input x gets closer to a specific point such as 0, ${\dfrac{\pi }{2}}$, or infinity. Even if the function is not explicitly defined at that point, the limit helps us understand its behavior. 

${\lim _{x\rightarrow a}f\left( x\right) =L}$

Formulas For Solving

To evaluate the limits of trigonometric functions at a specific point, say x = a, or as x approaches infinity (x → ±∞), we use the formulas given below.

For Limits at a Point

For a given point x = a, the limits of the trigonometric functions are:

• ${\lim_{x\to a} \sin x=\sin a}$
• ${\lim_{x\to a} \cos x=\cos a}$
• ${\lim_{x\to a} \tan x=\tan a}$
• ${\lim_{x\to a} \sec x=\sec a}$
• ${\lim_{x\to a} \text{cosec}\, x=\text{cosec}\, a}$
• ${\lim_{x\to a} \cot x=\cot a}$

These limits give finite values except when the function is undefined, such as when x = a, such as tan x or sec x at x = ${\dfrac{\pi }{2}+n\pi}$. 

For Limits at Infinity

As x approaches infinity x → ±∞, the trigonometric functions sin⁡x, cos⁡x, tan⁡x, etc., oscillate and do not give a single value. 

• ${\lim_{x\to \pm \infty} \sin x}$ = not defined
• ${\lim_{x\to \pm \infty} \cos x}$ = not defined
• ${\lim_{x\to \pm \infty} \tan x}$ = not defined
• ${\lim_{x\to \pm \infty} \sec x}$ = not defined
• ${\lim_{x\to \pm \infty} \text{cosec}\, x}$ = not defined
• ${\lim_{x\to \pm \infty} \cot x}$ = not defined

Solving Limits

Apart from the formulas, we use the given theorems and rules to evaluate the limits of the trigonometric functions.

By Theorem 1

If f and g are two real-valued functions defined in the same domain such that f(x) ≤ g(x) for all x in the domain approaching a, then

${\lim_{x\to a} f\left( x\right) =f\left( a\right)}$ and ${\lim_{x\to a} g\left( x\right) =g\left( a\right)}$

If both limits exist, then

${\lim_{x\to a} f\left( x\right) \leq g\left( x\right)}$

Now, let us find the limit of the function ${\lim _{x\rightarrow 0}\dfrac{\sin x}{x+1}}$

Step 1: Bounding the Function

Since -1 ≤ sin x ≤ 1,

⇒ ${-\dfrac{1}{x+1}\leq \dfrac{\sin x}{x+1}\leq \dfrac{1}{x+1}}$

Step 2: Evaluating the Bounds

As x → 0, 

${\lim _{x\rightarrow 0}-\dfrac{1}{x+1}=-1}$ and ${\lim _{x\rightarrow 0}\dfrac{1}{x+1}=1}$

Step 3: Applying Theorem 1

Since the bounds converge to 0, by Theorem 1, 

${\lim _{x\rightarrow 0}\dfrac{\sin x}{x+1}=0}$

By the Squeeze Theorem

If f, g, and h are real-valued functions with the same domains such that f(x) ≤ g(x) ≤ h(x) for all x in the domain approaching a, then

${\lim_{x\to a} f\left( x\right) =l}$ and ${\lim_{x\to a} g\left( x\right) =l}$

⇒ ${\lim_{x\to a} h\left( x\right) =l}$

Now, let us evaluate ${\lim_{x\to 0} x^{2}\sin \left( \dfrac{1}{x}\right)}$ 

Here, we will use the Squeeze theorem, also known as the Sandwich theorem.

Step 1: Bounding the Function

As we know, the sine function is bounded: ${-1\leq \sin \left( \dfrac{1}{x}\right) \leq 1}$

Step 2: Multiplying by x² 

Since x² ≥ 0, for all x

Now, on multiplying through by x², 

${-x^{2}\leq x^{2}\sin \left( \dfrac{1}{x}\right) \leq x^{2}}$ …..(i)

Step 3: Taking Limits of the Bounds

As x → 0, both -x² and x² approach 0. This implies:

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

Step 4: Applying the Squeeze theorem

Since (i) and (ii) satisfy, then by applying the squeeze theorem, we get

${\lim_{x\to 0} x^{2}\sin \left( \dfrac{1}{x}\right) =0}$

By L’Hôpital’s Rule

L’Hôpital’s Rule is used to evaluate limits that result in indeterminate forms, such as ${\dfrac{0}{0}}$ or ${\dfrac{\infty }{\infty }}$

According to the theorem, if a limit results in an indeterminate form, and the numerator f(x) and denominator g(x) are differentiable near the point of interest (except possibly at the point itself), then:

${\dfrac{f\left( x\right) }{g\left( x\right) }=\dfrac{f’\left( x\right) }{g’\left( x\right) }}$

Now, let us evaluate ${\lim_{x\to 0} \dfrac{\sin 2x}{x}}$

Step 1: Verifying Whether L’Hôpital’s Rule Can be Applied

Here, if we substitute x = 0 into ${\dfrac{\sin 2x}{x}}$ gives ${\dfrac{0}{0}}$, which is indeterminate.

Thus, L’Hôpital’s Rule can be applied.

Step 2: Applying L’Hôpital’s Rule 

Now, we will apply L’Hôpital’s Rule.

Differentiating the numerator and denominator with respect to x, we get

${\dfrac{d}{dx}\left( \sin 2x\right)}$ = 2 cos 2x

${\dfrac{d}{dx}\left( x\right)}$ = 1

Step 3: Simplifying the Limit

Substituting the derivations back into the limit, we get

${\lim_{x\to 0} \dfrac{\sin 2x}{x}}$

= ${\lim_{x\to 0} \dfrac{2\cos 2x}{1}}$ …..(i)

Step 4: Evaluating the Limit

As x → 0, cos 2x → cos 0 = 1

Now, from (i), we get

= ${\dfrac{2\times 1}{1}}$

= ${2}$