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Limits

Limit of Trigonometric Function

Understanding Limits of Trigonometric Functions

Imagine you observe a clock pendulum swinging very slowly approaching its equilibrium point. This motion is similar to the behavior of trigonometric functions when their variable approaches a certain value. Limits of trigonometric functions examine how the values of sine, cosine, and tangent functions behave when the input approaches critical points.

Unlike limits of algebraic functions that can often be solved by direct substitution, trigonometric functions have special characteristics due to their periodic and oscillating nature. This makes us need to use theorems and special properties to solve trigonometric limits.

Fundamental Theorem of Trigonometric Limits

The most important foundation in trigonometric limits is the theorem stating that the sine function approaches its gradient when the angle approaches zero.

Basic Sine Limit

The most fundamental theorem is:

lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1x→0lim​xsinx​=1

This theorem cannot be proven using direct substitution because it results in the form 00\frac{0}{0}00​. Its proof requires a geometric approach using the unit circle and the squeeze theorem.

In this theorem, xxx must be in radians, not degrees. If using degrees, the result will be different.

Consequences of the Basic Limit

From the fundamental limit above, we can derive several other important limits:

lim⁡x→0tan⁡xx=1(because tan⁡x=sin⁡xcos⁡x and cos⁡x→1)\lim_{x \to 0} \frac{\tan x}{x} = 1 \quad \text{(because } \tan x = \frac{\sin x}{\cos x} \text{ and } \cos x \to 1\text{)}x→0lim​xtanx​=1(because tanx=cosxsinx​ and cosx→1)
lim⁡x→01−cos⁡xx=0(use L’Hoˆpital or Taylor expansion)\lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \quad \text{(use L'Hôpital or Taylor expansion)}x→0lim​x1−cosx​=0(use L’Hoˆpital or Taylor expansion)
lim⁡x→01−cos⁡xx2=12(identity 1−cos⁡x=2sin⁡2x2)\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \quad \text{(identity } 1 - \cos x = 2\sin^2\frac{x}{2}\text{)}x→0lim​x21−cosx​=21​(identity 1−cosx=2sin22x​)

Properties of Trigonometric Limits

Based on the fundamental theorem, we can build a series of very useful properties:

Trigonometric Ratios

For constants a≠0a \neq 0a=0 and b≠0b \neq 0b=0, and all following limits when x→0x \to 0x→0:

LimitResultNotes
lim⁡x→0sin⁡axbx\lim_{x \to 0} \frac{\sin ax}{bx}limx→0​bxsinax​ab\frac{a}{b}ba​Manipulation from basic theorem
lim⁡x→0tan⁡axbx\lim_{x \to 0} \frac{\tan ax}{bx}limx→0​bxtanax​ab\frac{a}{b}ba​Because tan⁡ax=sin⁡axcos⁡ax\tan ax = \frac{\sin ax}{\cos ax}tanax=cosaxsinax​
lim⁡x→0sin⁡axsin⁡bx\lim_{x \to 0} \frac{\sin ax}{\sin bx}limx→0​sinbxsinax​ab\frac{a}{b}ba​Combination of two sine limits
lim⁡x→0tan⁡axtan⁡bx\lim_{x \to 0} \frac{\tan ax}{\tan bx}limx→0​tanbxtanax​ab\frac{a}{b}ba​Combination of two tangent limits

Trigonometric Combinations

From the trigonometric ratio properties, we can derive several trigonometric combination properties:

lim⁡x→0sin⁡axsin⁡bx=ab\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}x→0lim​sinbxsinax​=ba​
lim⁡x→0tan⁡axsin⁡bx=ab\lim_{x \to 0} \frac{\tan ax}{\sin bx} = \frac{a}{b}x→0lim​sinbxtanax​=ba​

Techniques for Solving Trigonometric Limits

Substitution and Manipulation Techniques

When facing complex trigonometric limits, we often need to manipulate expressions to use the fundamental theorem.

Calculate: lim⁡x→0sin⁡2x3x\lim_{x \to 0} \frac{\sin 2x}{3x}limx→0​3xsin2x​

We manipulate to obtain the standard form:

lim⁡x→0sin⁡2x3x=lim⁡x→02sin⁡2x2⋅3x=lim⁡x→023⋅sin⁡2x2x\lim_{x \to 0} \frac{\sin 2x}{3x} = \lim_{x \to 0} \frac{2 \sin 2x}{2 \cdot 3x} = \lim_{x \to 0} \frac{2}{3} \cdot \frac{\sin 2x}{2x}x→0lim​3xsin2x​=x→0lim​2⋅3x2sin2x​=x→0lim​32​⋅2xsin2x​
=23lim⁡x→0sin⁡2x2x=23⋅1=23= \frac{2}{3} \lim_{x \to 0} \frac{\sin 2x}{2x} = \frac{2}{3} \cdot 1 = \frac{2}{3}=32​x→0lim​2xsin2x​=32​⋅1=32​

Trigonometric Identity Techniques

Often we need to use trigonometric identities to simplify expressions. Always ensure the function is defined at the point being approached.

Calculate: lim⁡x→π4sin⁡x+cos⁡xsin⁡x\lim_{x \to \frac{\pi}{4}} \frac{\sin x + \cos x}{\sin x}limx→4π​​sinxsinx+cosx​

Since sin⁡(π4)=22≠0\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \neq 0sin(4π​)=22​​=0, the denominator is not zero at x=π4x = \frac{\pi}{4}x=4π​.

Direct substitution:

lim⁡x→π4sin⁡x+cos⁡xsin⁡x=sin⁡π4+cos⁡π4sin⁡π4\lim_{x \to \frac{\pi}{4}} \frac{\sin x + \cos x}{\sin x} = \frac{\sin \frac{\pi}{4} + \cos \frac{\pi}{4}}{\sin \frac{\pi}{4}}x→4π​lim​sinxsinx+cosx​=sin4π​sin4π​+cos4π​​
=22+2222=2⋅2222=222=2⋅22=2= \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{\frac{\sqrt{2}}{2}} = \sqrt{2} \cdot \frac{2}{\sqrt{2}} = 2=22​​22​​+22​​​=22​​2⋅22​​​=22​​2​​=2​⋅2​2​=2

Techniques for Special Forms

For limits involving the form 00\frac{0}{0}00​, we need special techniques.

Calculate: lim⁡x→π2sin⁡2(x−π2)x(x−π2)\lim_{x \to \frac{\pi}{2}} \frac{\sin^2(x - \frac{\pi}{2})}{x(x - \frac{\pi}{2})}limx→2π​​x(x−2π​)sin2(x−2π​)​

Let u=x−π2u = x - \frac{\pi}{2}u=x−2π​, then when x→π2x \to \frac{\pi}{2}x→2π​, we have u→0u \to 0u→0 and x=u+π2x = u + \frac{\pi}{2}x=u+2π​.

lim⁡u→0sin⁡2u(u+π2)⋅u=lim⁡u→0sin⁡2uu2⋅uu+π2\lim_{u \to 0} \frac{\sin^2 u}{(u + \frac{\pi}{2}) \cdot u} = \lim_{u \to 0} \frac{\sin^2 u}{u^2} \cdot \frac{u}{u + \frac{\pi}{2}}u→0lim​(u+2π​)⋅usin2u​=u→0lim​u2sin2u​⋅u+2π​u​
=lim⁡u→0(sin⁡uu)2⋅lim⁡u→0uu+π2= \lim_{u \to 0} \left(\frac{\sin u}{u}\right)^2 \cdot \lim_{u \to 0} \frac{u}{u + \frac{\pi}{2}}=u→0lim​(usinu​)2⋅u→0lim​u+2π​u​
=12⋅lim⁡u→0uπ2+u=1⋅0π2=0= 1^2 \cdot \lim_{u \to 0} \frac{u}{\frac{\pi}{2} + u} = 1 \cdot \frac{0}{\frac{\pi}{2}} = 0=12⋅u→0lim​2π​+uu​=1⋅2π​0​=0

Trigonometric Limits with Angle Identities

Using Sum and Difference Formulas

When dealing with trigonometric functions involving sum or difference of angles, we can use trigonometric identities.

Calculate: lim⁡t→0cot⁡5tcot⁡10t\lim_{t \to 0} \frac{\cot 5t}{\cot 10t}limt→0​cot10tcot5t​

Using the definition of cotangent:

lim⁡t→0cot⁡5tcot⁡10t=lim⁡t→0cos⁡5tsin⁡5tcos⁡10tsin⁡10t=lim⁡t→0cos⁡5tsin⁡10tsin⁡5tcos⁡10t\lim_{t \to 0} \frac{\cot 5t}{\cot 10t} = \lim_{t \to 0} \frac{\frac{\cos 5t}{\sin 5t}}{\frac{\cos 10t}{\sin 10t}} = \lim_{t \to 0} \frac{\cos 5t \sin 10t}{\sin 5t \cos 10t}t→0lim​cot10tcot5t​=t→0lim​sin10tcos10t​sin5tcos5t​​=t→0lim​sin5tcos10tcos5tsin10t​
=lim⁡t→0cos⁡5tcos⁡10t⋅sin⁡10tsin⁡5t=1⋅105=2= \lim_{t \to 0} \frac{\cos 5t}{\cos 10t} \cdot \frac{\sin 10t}{\sin 5t} = 1 \cdot \frac{10}{5} = 2=t→0lim​cos10tcos5t​⋅sin5tsin10t​=1⋅510​=2

Exercises

  1. Calculate lim⁡x→0sin⁡3x4x\lim_{x \to 0} \frac{\sin 3x}{4x}limx→0​4xsin3x​

  2. Calculate lim⁡x→0tan⁡2xsin⁡5x\lim_{x \to 0} \frac{\tan 2x}{\sin 5x}limx→0​sin5xtan2x​

  3. Calculate lim⁡x→01−cos⁡3xx2\lim_{x \to 0} \frac{1 - \cos 3x}{x^2}limx→0​x21−cos3x​

  4. Calculate lim⁡x→π6sin⁡x−12x−π6\lim_{x \to \frac{\pi}{6}} \frac{\sin x - \frac{1}{2}}{x - \frac{\pi}{6}}limx→6π​​x−6π​sinx−21​​

  5. Calculate lim⁡x→0sin⁡xcos⁡xx\lim_{x \to 0} \frac{\sin x \cos x}{x}limx→0​xsinxcosx​

Answer Key

  1. Solution:

    Use algebraic manipulation to obtain the standard form:

    lim⁡x→0sin⁡3x4x=lim⁡x→03sin⁡3x3⋅4x=lim⁡x→034⋅sin⁡3x3x\lim_{x \to 0} \frac{\sin 3x}{4x} = \lim_{x \to 0} \frac{3 \sin 3x}{3 \cdot 4x} = \lim_{x \to 0} \frac{3}{4} \cdot \frac{\sin 3x}{3x}x→0lim​4xsin3x​=x→0lim​3⋅4x3sin3x​=x→0lim​43​⋅3xsin3x​
    =34lim⁡x→0sin⁡3x3x=34⋅1=34= \frac{3}{4} \lim_{x \to 0} \frac{\sin 3x}{3x} = \frac{3}{4} \cdot 1 = \frac{3}{4}=43​x→0lim​3xsin3x​=43​⋅1=43​

  2. Solution:

    Use trigonometric ratio properties:

    lim⁡x→0tan⁡2xsin⁡5x=lim⁡x→0tan⁡2x2x⋅5xsin⁡5x⋅2x5x\lim_{x \to 0} \frac{\tan 2x}{\sin 5x} = \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \frac{5x}{\sin 5x} \cdot \frac{2x}{5x}x→0lim​sin5xtan2x​=x→0lim​2xtan2x​⋅sin5x5x​⋅5x2x​
    =lim⁡x→0tan⁡2x2x⋅lim⁡x→05xsin⁡5x⋅25= \lim_{x \to 0} \frac{\tan 2x}{2x} \cdot \lim_{x \to 0} \frac{5x}{\sin 5x} \cdot \frac{2}{5}=x→0lim​2xtan2x​⋅x→0lim​sin5x5x​⋅52​
    =1⋅1⋅25=25= 1 \cdot 1 \cdot \frac{2}{5} = \frac{2}{5}=1⋅1⋅52​=52​

  3. Solution:

    Use the identity 1−cos⁡ax=2sin⁡2ax21 - \cos ax = 2\sin^2\frac{ax}{2}1−cosax=2sin22ax​:

    lim⁡x→01−cos⁡3xx2=lim⁡x→02sin⁡23x2x2\lim_{x \to 0} \frac{1 - \cos 3x}{x^2} = \lim_{x \to 0} \frac{2\sin^2\frac{3x}{2}}{x^2}x→0lim​x21−cos3x​=x→0lim​x22sin223x​​
    =lim⁡x→02⋅(sin⁡3x23x2)2⋅(32)2= \lim_{x \to 0} 2 \cdot \left(\frac{\sin\frac{3x}{2}}{\frac{3x}{2}}\right)^2 \cdot \left(\frac{3}{2}\right)^2=x→0lim​2⋅(23x​sin23x​​)2⋅(23​)2
    =2⋅12⋅94=92= 2 \cdot 1^2 \cdot \frac{9}{4} = \frac{9}{2}=2⋅12⋅49​=29​

  4. Solution:

    Let h=x−π6h = x - \frac{\pi}{6}h=x−6π​, then x=h+π6x = h + \frac{\pi}{6}x=h+6π​ and when x→π6x \to \frac{\pi}{6}x→6π​, we have h→0h \to 0h→0:

    lim⁡h→0sin⁡(h+π6)−12h\lim_{h \to 0} \frac{\sin(h + \frac{\pi}{6}) - \frac{1}{2}}{h}h→0lim​hsin(h+6π​)−21​​
    =lim⁡h→0sin⁡hcos⁡π6+cos⁡hsin⁡π6−12h= \lim_{h \to 0} \frac{\sin h \cos\frac{\pi}{6} + \cos h \sin\frac{\pi}{6} - \frac{1}{2}}{h}=h→0lim​hsinhcos6π​+coshsin6π​−21​​
    =lim⁡h→0sin⁡h⋅32+cos⁡h⋅12−12h= \lim_{h \to 0} \frac{\sin h \cdot \frac{\sqrt{3}}{2} + \cos h \cdot \frac{1}{2} - \frac{1}{2}}{h}=h→0lim​hsinh⋅23​​+cosh⋅21​−21​​
    =lim⁡h→032sin⁡h+12(cos⁡h−1)h= \lim_{h \to 0} \frac{\frac{\sqrt{3}}{2}\sin h + \frac{1}{2}(\cos h - 1)}{h}=h→0lim​h23​​sinh+21​(cosh−1)​
    =32lim⁡h→0sin⁡hh+12lim⁡h→0cos⁡h−1h= \frac{\sqrt{3}}{2} \lim_{h \to 0} \frac{\sin h}{h} + \frac{1}{2} \lim_{h \to 0} \frac{\cos h - 1}{h}=23​​h→0lim​hsinh​+21​h→0lim​hcosh−1​
    =32⋅1+12⋅0=32= \frac{\sqrt{3}}{2} \cdot 1 + \frac{1}{2} \cdot 0 = \frac{\sqrt{3}}{2}=23​​⋅1+21​⋅0=23​​

  5. Solution:

    Use the identity sin⁡xcos⁡x=12sin⁡2x\sin x \cos x = \frac{1}{2}\sin 2xsinxcosx=21​sin2x:

    lim⁡x→0sin⁡xcos⁡xx=lim⁡x→012sin⁡2xx\lim_{x \to 0} \frac{\sin x \cos x}{x} = \lim_{x \to 0} \frac{\frac{1}{2}\sin 2x}{x}x→0lim​xsinxcosx​=x→0lim​x21​sin2x​
    =12lim⁡x→0sin⁡2xx=12lim⁡x→02sin⁡2x2x= \frac{1}{2} \lim_{x \to 0} \frac{\sin 2x}{x} = \frac{1}{2} \lim_{x \to 0} \frac{2 \sin 2x}{2x}=21​x→0lim​xsin2x​=21​x→0lim​2x2sin2x​
    =12⋅2lim⁡x→0sin⁡2x2x=1⋅1=1= \frac{1}{2} \cdot 2 \lim_{x \to 0} \frac{\sin 2x}{2x} = 1 \cdot 1 = 1=21​⋅2x→0lim​2xsin2x​=1⋅1=1
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Limit of Algebraic Function

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Application of Limit Function

  • Limit of Trigonometric FunctionMaster trigonometric limits using fundamental theorems. Learn sin x/x equals 1, ratio properties, identities, and advanced substitution techniques.
On this page
  • Understanding Limits of Trigonometric Functions
  • Fundamental Theorem of Trigonometric Limits
    • Basic Sine Limit
    • Consequences of the Basic Limit
  • Properties of Trigonometric Limits
    • Trigonometric Ratios
    • Trigonometric Combinations
  • Techniques for Solving Trigonometric Limits
    • Substitution and Manipulation Techniques
    • Trigonometric Identity Techniques
    • Techniques for Special Forms
  • Trigonometric Limits with Angle Identities
    • Using Sum and Difference Formulas
  • Exercises
    • Answer Key
  • Comments
  • Report
  • Source code