Find $k$ such that $g(x)$ is continuous at $x=0$

Question

Find a nonzero $k$ such that
$$g(x) = \begin{cases} \frac{\tan kx}{x}, & \text{if $x$ $\lt$ 0} \\[2ex] 3x + 2k^2, & \text{if $x$ $\ge$ 0} \end{cases}$$ is continuous at $x=0$.

EDIT: I attempted to tackle this but only got $k=0$ as an answer because I thought $\lim_x\to0\frac{\tan kx}{x}=0$. However, the previous limit happens to be a fundamental one whose correct value is $1%$ rather than $0$.

Thanks in advance.

Answer 1

You want to find such that the right limit and the left limit at $x=0$ are the same.

$$g(x) = \begin{cases} \frac{\tan kx}{x}, & \text{if $x$ $\lt$ 0} \\[2ex] 3x + 2k^2, & \text{if $x$ $\ge$ 0} \end{cases}$$

Note that $$ \frac{\tan kx}{x} = \frac{\sin(kx)}{x\cdot\cos(kx)} = \frac{k}{\cos kx}\cdot\frac{\sin(kx)}{kx} $$ so the left limit is $k$. Meanwhile the right limit is $2k^2$. So for continuity we solve $$2k^2=k$$ which gives $ k=0$ or $k=1/2$ .

Answer 2

As $x\to 0$, $\frac{\tan(kx)}{x} = \frac{\sin(kx)}{x \cos(kx)} = k \cdot \frac{\sin(kx)}{kx } \cdot \frac{1}{\cos(kx)} \to k$, and $3x+2k^2\to 2k^2$. For continuity $k=2k^2$ or $k=\frac{1}{2}$ or $k=0$.

Answer 3

Maybe， you are just confused by the limit $$\lim_{x \to 0}\frac{\tan kx}{x}.$$

To evaluate it, you should notice that $$\tan x \sim x ,x \to 0.$$ Thus, substitute $x$ inside for $kx$. We obtain $$\tan kx \sim kx, kx \to 0.$$but $kx \to 0$ if and only if $x \to 0.$ Therefore, $$\tan kx \sim kx,x \to 0.$$ It follows that $$\lim_{x \to 0}\frac{\tan kx}{x}=\lim_{x \to 0}\left(k \cdot\frac{\tan kx}{kx}\right)=k\cdot \lim_{x \to 0}\frac{\tan kx}{kx}=k\cdot 1=k.$$