What value should be assigned to k so that h will be continous?

Question

Consider the function $h(x)$ defined by $$ h(x) = \begin{cases} \frac{\tan 5x}{4x} & x \neq 0 \\ k & x = 0 \end{cases} $$ What value should be assigned to $k$ to assure that $h$ is continuous?

My approach: I tried to find a way to rewrite the function (tan 5x)/ 4x.

1. That can be rewritten as ((sin 5x)/ (cos 5x))/ 4x
2. ((sin 5x)/ (cos 5x)) * 1 / 4x
3. I kept in mind that there is a rule that sin x / x is equal to 1
4. I rewrote the expression to be (sin 5x / x) * (1 / (cos 5x) * 4)
5. 1 * (1 / (4 cos 5x)); There can be no value of the function where x = 0
6. The graph shows that as well: [1]: https://i.stack.imgur.com/KgM6t.png
7. At a value between 1.2 and 1.3 lies the value of y which must be defined in ordered to make the function continuous.
8. But I cannot solve this by graphing it- it must be done by hand. So how will I go from step 5 to getting the value of k that will make this function h, continuous?

Answer 1

Your idea is good but there are many issues as indicated in the comments.

Simply, to assure continuity, by the definition, we need to assume:

$$k=\lim_{x\to 0} \frac{\tan (5x)}{4x}=\lim_{x\to 0} \left(\frac54\cdot \frac{\sin (5x)}{5x}\cdot\frac{1}{\cos(5x)}\right)=\frac 54$$

in such way that

$$h(0)=\lim_{x\to 0} h(x)$$

Answer 2

By the definition of $h(x)$ provided by you, it can be can say that for $h(x)$ to be a continuous function $$\lim_{x\to 0}h(x)=h(0)$$

That is, $$\lim_{x\to 0} \frac{\tan(5x)}{4x}=k$$ (with $k\in \Bbb R$) which we can rewrite as, $$\lim_{x\to 0} \frac{\tan(5x)}{5x}\times\frac{5}{4}=k$$ Now since, $$\lim_{x\to 0} \frac{\tan 5x}{5x}=1$$

so, $$\frac{5}{4}=k=1.25.$$ Like you said, $k$ should be between 1.2 and 1.3. Do take a look at comments highlighting some of your mistakes regarding trigonometric limits.