Limit of sin(θ)/θ as θ tends to 0

The diagram at right shows a circle with centre O and radius r = 1. Let two radii OA and OB make an arc of θ radians. Since we are considering the limit as θ tends to zero, we may assume θ is a small positive number, say 0 < θ < ½ π in the first quadrant.

In the diagram, let R₁ be the triangle OAB, R₂ the circular sector OAB, and R₃ the triangle OAC.

The area of triangle OAB is:

${\displaystyle \mathrm {Area} (R_{1})={\tfrac {1}{2}}\ |OA|\ |OB|\sin \theta ={\tfrac {1}{2}}\sin \theta \,.}$

The area of the circular sector OAB is:

${\displaystyle \mathrm {Area} (R_{2})={\tfrac {1}{2}}\theta \,.}$

The area of the triangle OAC is given by:

${\displaystyle \mathrm {Area} (R_{3})={\tfrac {1}{2}}\ |OA|\ |AC|={\tfrac {1}{2}}\tan \theta \,.}$

Since each region is contained in the next, one has:

${\displaystyle {\text{Area}}(R_{1})<{\text{Area}}(R_{2})<{\text{Area}}(R_{3})\implies {\tfrac {1}{2}}\sin \theta <{\tfrac {1}{2}}\theta <{\tfrac {1}{2}}\tan \theta \,.}$

Moreover, since sin θ > 0 in the first quadrant, we may divide through by ½ sin θ, giving:

${\displaystyle 1<{\frac {\theta }{\sin \theta }}<{\frac {1}{\cos \theta }}\implies 1>{\frac {\sin \theta }{\theta }}>\cos \theta \,.}$

In the last step we took the reciprocals of the three positive terms, reversing the inequities.

We conclude that for 0 < θ < ½ π, the quantity sin(θ)/θ is always less than 1 and always greater than cos(θ). Thus, as θ gets closer to 0, sin(θ)/θ is "squeezed" between a ceiling at height 1 and a floor at height cos θ, which rises towards 1; hence sin(θ)/θ must tend to 1 as θ tends to 0 from the positive side:

${\displaystyle \lim _{\theta \to 0^{+}}{\frac {\sin \theta }{\theta }}=1\,.}$

For the case where θ is a small negative number –½ π < θ < 0, we use the fact that sine is an odd function:

${\displaystyle \lim _{\theta \to 0^{-}}\!{\frac {\sin \theta }{\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {\sin(-\theta )}{-\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {-\sin \theta }{-\theta }}\ =\ \lim _{\theta \to 0^{+}}\!{\frac {\sin \theta }{\theta }}\ =\ 1\,.}$

Limit of (cos(θ)-1)/θ as θ tends to 0

The last section enables us to calculate this new limit relatively easily. This is done by employing a simple trick. In this calculation, the sign of θ is unimportant.

${\displaystyle \lim _{\theta \to 0}\,{\frac {\cos \theta -1}{\theta }}\ =\ \lim _{\theta \to 0}\left({\frac {\cos \theta -1}{\theta }}\right)\!\!\left({\frac {\cos \theta +1}{\cos \theta +1}}\right)\ =\ \lim _{\theta \to 0}\,{\frac {\cos ^{2}\!\theta -1}{\theta \,(\cos \theta +1)}}.}$

Using cos²θ – 1 = –sin²θ, the fact that the limit of a product is the product of limits, and the limit result from the previous section, we find that:

${\displaystyle \lim _{\theta \to 0}\,{\frac {\cos \theta -1}{\theta }}\ =\ \lim _{\theta \to 0}\,{\frac {-\sin ^{2}\theta }{\theta (\cos \theta +1)}}\ =\ \left(-\lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}\right)\!\left(\lim _{\theta \to 0}\,{\frac {\sin \theta }{\cos \theta +1}}\right)\ =\ (-1)\left({\frac {0}{2}}\right)=0\,.}$

Limit of tan(θ)/θ as θ tends to 0

Using the limit for the sine function, the fact that the tangent function is odd, and the fact that the limit of a product is the product of limits, we find:

${\displaystyle \lim _{\theta \to 0}{\frac {\tan \theta }{\theta }}\ =\ \left(\lim _{\theta \to 0}{\frac {\sin \theta }{\theta }}\right)\!\left(\lim _{\theta \to 0}{\frac {1}{\cos \theta }}\right)\ =\ (1)(1)\ =\ 1\,.}$

Derivative of the sine function

We calculate the derivative of the sine function from the limit definition:

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =\lim _{\delta \to 0}{\frac {\sin(\theta +\delta )-\sin \theta }{\delta }}.}$

Using the angle addition formula sin(α+β) = sin α cos β + sin β cos α, we have:

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =\lim _{\delta \to 0}{\frac {\sin \theta \cos \delta +\sin \delta \cos \theta -\sin \theta }{\delta }}=\lim _{\delta \to 0}\left({\frac {\sin \delta }{\delta }}\cos \theta +{\frac {\cos \delta -1}{\delta }}\sin \theta \right).}$

Using the limits for the sine and cosine functions:

${\displaystyle {\frac {\operatorname {d} }{\operatorname {d} \!\theta }}\,\sin \theta =(1)\cos \theta +(0)\sin \theta =\cos \theta \,.}$

Derivative of the cosine function

Derivative of the tangent function

Differentiating the inverse sine function

We let

${\displaystyle y=\arcsin x\,\!}$

Where

${\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}$

Then

${\displaystyle \sin y=x\,\!}$

Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:

${\displaystyle {d \over dx}\sin y={d \over dx}x}$

${\displaystyle \cos y\cdot {dy \over dx}=1\,\!}$

Substituting ${\displaystyle \cos y={\sqrt {1-\sin ^{2}y}}}$ in from above,

${\displaystyle {\sqrt {1-\sin ^{2}y}}\cdot {dy \over dx}=1}$

Substituting ${\displaystyle x=\sin y}$ in from above,

${\displaystyle {\sqrt {1-x^{2}}}\cdot {dy \over dx}=1}$

${\displaystyle {dy \over dx}={\frac {1}{\sqrt {1-x^{2}}}}}$

Differentiating the inverse cosine function

We let

${\displaystyle y=\arccos x\,\!}$

Where

${\displaystyle 0\leq y\leq \pi }$

Then

${\displaystyle \cos y=x\,\!}$

Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:

${\displaystyle {d \over dx}\cos y={d \over dx}x}$

${\displaystyle -\sin y\cdot {dy \over dx}=1}$

Substituting ${\displaystyle \sin y={\sqrt {1-\cos ^{2}y}}\,\!}$ in from above, we get

${\displaystyle -{\sqrt {1-\cos ^{2}y}}\cdot {dy \over dx}=1}$

Substituting ${\displaystyle x=\cos y\,\!}$ in from above, we get

${\displaystyle -{\sqrt {1-x^{2}}}\cdot {dy \over dx}=1}$

${\displaystyle {dy \over dx}=-{\frac {1}{\sqrt {1-x^{2}}}}}$

Alternatively, once the derivative of ${\displaystyle \arcsin x}$ is established, the derivative of ${\displaystyle \arccos x}$ follows immediately by differentiating the identity ${\displaystyle \arcsin x+\arccos x=\pi /2}$ so that ${\displaystyle (\arccos x)'=-(\arcsin x)'}$.

Differentiating the inverse tangent function

We let

${\displaystyle y=\arctan x\,\!}$

Where

${\displaystyle -{\frac {\pi }{2}}<y<{\frac {\pi }{2}}}$

Then

${\displaystyle \tan y=x\,\!}$

Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:

${\displaystyle {d \over dx}\tan y={d \over dx}x}$

Left side:

${\displaystyle {d \over dx}\tan y=\sec ^{2}y\cdot {dy \over dx}=(1+\tan ^{2}y){dy \over dx}}$ using the Pythagorean identity

Right side:

${\displaystyle {d \over dx}x=1}$

Therefore,

${\displaystyle (1+\tan ^{2}y){dy \over dx}=1}$

Substituting ${\displaystyle x=\tan y\,\!}$ in from above, we get

${\displaystyle (1+x^{2}){dy \over dx}=1}$

${\displaystyle {dy \over dx}={\frac {1}{1+x^{2}}}}$

Differentiating the inverse cotangent function

We let

${\displaystyle y=\operatorname {arccot} x}$

where ${\displaystyle 0<y<\pi }$. Then

${\displaystyle \cot y=x}$

Taking the derivative with respect to ${\displaystyle x}$ on both sides and solving for dy/dx:

${\displaystyle {\frac {d}{dx}}\cot y={\frac {d}{dx}}x}$

Left side:

${\displaystyle {d \over dx}\cot y=-\csc ^{2}y\cdot {dy \over dx}=-(1+\cot ^{2}y){dy \over dx}}$ using the Pythagorean identity

Right side:

${\displaystyle {d \over dx}x=1}$

Therefore,

${\displaystyle -(1+\cot ^{2}y){\frac {dy}{dx}}=1}$

Substituting ${\displaystyle x=\cot y}$,

${\displaystyle -(1+x^{2}){\frac {dy}{dx}}=1}$

${\displaystyle {\frac {dy}{dx}}=-{\frac {1}{1+x^{2}}}}$

Alternatively, as the derivative of ${\displaystyle \arctan x}$ is derived as shown above, then using the identity ${\displaystyle \arctan x+\operatorname {arccot} x={\dfrac {\pi }{2}}}$ follows immediately that

$${\displaystyle {\begin{aligned}{\dfrac {d}{dx}}\operatorname {arccot} x&={\dfrac {d}{dx}}\left({\dfrac {\pi }{2}}-\arctan x\right)\\&=-{\dfrac {1}{1+x^{2}}}\end{aligned}}}$$

Differentiating the inverse secant function

Differentiating the inverse cosecant function