I have this piecewise function
$f(x) = \begin{cases} \frac{\sin(-8x)}{8x} & x < 0
\\ (2x + 9k – 7) & x \ge 0 \end{cases}$
Now the formula for calculating the limit of $\frac{\sin(x)}{x}$ is
$\frac{\sin(x)}{x} = 1$
So in this case, you would divide $-8$ by $8$, to show that the limit of the first function is $-1$
However, I have a problem with this.
To solve if two functions are continuous, you have to determine that the limit of the first function is equal to the limit of the second function. This is typically done, by setting the limit equal to where they would converge, in this case at $0$.
Here is the formula
$\lim \limits_{x \to a- } f(x)= \lim \limits_{x \to a+ } g(x)$
This would make the denominator for the first function $0$, and you can't divide by $0$.
So why does the first function evaluate to $-1$ instead of undefined.
Best Answer
So your question basically boils down to:
For that answer, I am putting this answer of mine from this question:
From the Maclaurin series expansion of $f(x)=\sin x$,we have that $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$
or, considering $x\not = 0,$ $$\frac{\sin x}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots$$ And hence by using the notion of uniform convergence of a power series, it can be said that $$\lim_\limits{x\to 0} \frac{\sin x}{x} = 1$$
Thus at $x=0$, the function $f(x)=\frac{\sin x}{x}$ has R.H.L.=L.H.L.=$1 \not = f(0)$
So at $x=0$, $f(x)$ has removable discontinuity.
Hence to make the function continuous at $x=0$, $f(0)$ is defined to be $1$.
So the actual function should be like $$f(x)=\begin{cases}\frac{\sin x}{x} & x\not = 0 \\ \, \, \, 1 & x=0\end{cases}$$