## Content

### 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*}

Exercise 6

Find

1. $$\displaystyle \lim_{x\to 1} \dfrac{\sqrt{x^2+15}-4}{x-1}$$
2. $$\displaystyle \lim_{x\to 4} \dfrac{\dfrac{1}{x}-\dfrac{1}{4}}{x-4}$$.

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.

Next page - Content - Continuity