Trigonometric Limits

\lim\limits_{x \to 0}\frac{sin(x)}{x}=?

Unit Circle

Remember cos(x)=\frac{OA}{OB}=\frac{OA}{1}, hence OA=cos(x) and the co-ordinate of A is (cos(x), 0).

sin(x)=\frac{AB}{OB}=\frac{AB}{1}, hence AB=sin(x) and the co-ordinate of B is (cos(x), sin(x))

And from the definition of tan(x) we know D is the point (1, tan(x))

Consider the areas of triangle OAB, sector OBC, and triangle OCD.

We know from inspection of the above diagram that

Area OAB< Area OCB<Area OCD

Which means,

\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2

We can ignore all of the halves.

cos(x)sin(x)<x<(1)tan(x)

Remember tan(x)=\frac{sin(x)}{cos(x)}

cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}

Divide everything by sin(x) (as we are in the first quadrant we know sin(x)>0, so we don’t need to worry about the inequality)

cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}

Invert everything and change the direction of the inequalities)

\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)

I am going to rewrite it as follows

cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}

because I like to use less thans rather than greater thans.

Now what happens as x tends to 0?

cos(0)=1

1<\frac{sin(x)}{x}<\frac{1}{1}

Hence by the squeeze theorem \lim\limits_{x \to 0}\frac{sin(x)}{x}=1

Now we know this limit, we are going to use it to find \lim\limits_{x \to 0}\frac{1-cos(x)}{x}

Multiply by \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}

\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}

\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)

If we evaluate the limits,

(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0

Hence, \lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0

In the next post we are going to use these limits to differentiate sine and cosine functions.

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Filed under Area, Area of Triangles (Sine), Calculus, Identities, Trigonometry, Year 12 Mathematical Methods

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  1. Pingback: Differentiating Trigonometric Functions | Racquel Sanderson

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