How to solve trigonometric limit problems

Formulas we use to solve limit problems

Trigonometric limit problems revolve around three formulas:

$\lim_{\theta\to0}\sin{\theta}=0$

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

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

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When we solve trigonometric limit problems, our goal is always to reduce the function to a combination of nothing but these three formulas and simple constants.

How to solve trigonometric limit problems with trig limit formulas

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Let’s do an example where we evaluate the trig limit

Example

Evaluate the limit.

$\lim_{x\to0}\frac{1-\cos{x}}{x}$

Since we have exactly two terms in the numerator, we’re actually going to borrow the conjugate method for the first step of this problem.

$\lim_{x\to0}\frac{1-\cos{x}}{x}\cdot\left(\frac{1+\cos{x}}{1+\cos{x}}\right)$

$\lim_{x\to0}\frac{1-\cos^2{x}}{x(1+\cos{x})}$

Applying the identity $1-\cos^2{x}=\sin^2{x}$ to the numerator gives

$\lim_{x\to0}\frac{\sin^2{x}}{x(1+\cos{x})}$

When we solve trigonometric limit problems, our goal is always to reduce the function to a combination of nothing but these three formulas and simple constants.

Notice now that we can factor out $(\sin{x})/x$, which is one of our three fundamental formulas.

$\lim_{x\to0}\frac{\sin{x}}{x}\cdot\frac{\sin{x}}{1+\cos{x}}$

$\lim_{x\to0}\frac{\sin{x}}{x}\cdot\lim_{x\to0}\frac{\sin{x}}{1+\cos{x}}$

Since the first limit is one of our three fundamental formulas, we can replace it with its value.

$1\cdot\lim_{x\to0}\frac{\sin{x}}{1+\cos{x}}$

$\lim_{x\to0}\frac{\sin{x}}{1+\cos{x}}$

And now we can evaluate the limit at $x\to 0$ without making the denominator $0$.

$\frac{\sin{0}}{1+\cos{0}}$

$\frac{0}{1+1}$

$0$

So the limit is $0$.

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

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