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Further examples

There are some examples of limits that require some `tricks'.

For example, consider the limit

\[ \lim_{x\to 0} \dfrac{\sqrt{x^2+4}-2}{x^2}. \]

We cannot substitute \(x=0\), since then the denominator will be 0. To find this limit, we need to rationalise the numerator:

\begin{align*} \lim_{x\to 0} \dfrac{\sqrt{x^2+4}-2}{x^2} &= \lim_{x\to 0} \Biggl(\dfrac{\sqrt{x^2+4}-2}{x^2} \times \dfrac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}+2}\Biggr)\\ &= \lim_{x\to 0} \dfrac{(x^2+4)-4}{x^2\bigl(\sqrt{x^2+4}+2\bigr)}\\ &= \lim_{x\to 0}\dfrac{1}{\sqrt{x^2+4}+2} = \dfrac{1}{4}. \end{align*}

So far in this module, we have implicitly assumed the following facts — none of which we can prove without a more formal definition of limit.

Algebra of limits

Suppose that \(f(x)\) and \(g(x)\) are functions and that \(a\) and \(k\) are real numbers. If both \(\lim\limits_{x\to a} f(x)\) and \(\lim\limits_{x\to a} g(x)\) exist, then

1. \(\displaystyle \lim_{x\to a} \bigl(f(x) + g(x)\bigr) = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)\)
2. \(\displaystyle \lim_{x\to a} kf(x) = k\lim_{x\to a} f(x)\)
3. \(\displaystyle \lim_{x\to a} f(x)g(x) = \Bigl(\lim_{x\to a} f(x) \Bigr) \Bigl(\lim_{x\to a} g(x) \Bigr)\)
4. \(\displaystyle \lim_{x\to a} \dfrac{f(x)}{g(x)} = \dfrac{\lim\limits_{x\to a} f(x)}{\lim\limits_{x\to a} g(x)}\), provided \(\lim\limits_{x\to a} {g(x)}\) is not equal to 0.

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