Find the constant k that will make this piecewise continuous.

[kaderyo94]

Homework Statement

Find a value for the constant k that will make the function below continuous:

f(x)=\frac{x-1}{x^2-1}\ \text{if}\ x<=0
f(x)=\frac{tankx}{2x}~\text{if}~x>0

Homework Equations

The Attempt at a Solution

I've tried the only solution I can think of, which is to make
\frac{x-1}{x^2-1} = \frac{tankx}{2x}

And then I plug in 0 to try and get k, but I end up with 1 = 0/0. I know you are supposed to do something to the second equation to remedy this, but I cannot figure out what. I am fairly new to Calculus, so some help would be greatly appreciated.

 

[SammyS]

Homework Statement

Find a value for the constant k that will make the function below continuous:

f(x)=\frac{x-1}{x^2-1}\ \text{if}\ x<=0

f(x)=\frac{\tan kx}{2x}~\text{if}~x>0

Homework Equations

The Attempt at a Solution

I've tried the only solution I can think of, which is to make
\frac{x-1}{x^2-1} = \frac{\tan kx}{2x}

And then I plug in 0 to try and get k, but I end up with 1 = 0/0. I know you are supposed to do something to the second equation to remedy this, but I cannot figure out what. I am fairly new to Calculus, so some help would be greatly appreciated.

Hello kaderyo94. Welcome to PF !

Each piece of this piecewise-defined function has one or more discontinuities in its portion of the domain of the overall function. You can't "fix" those discontinuities by a choice of k.

I suspect the problem is: Find a value for the constant k that will make the function continuous at x = 0, which is where the two "pieces" join.

If that's the problem to be solved, then:

What must be true for the following limit to exist?

\displaystyle \lim_{x\to\,0}\,f(x)​

Then, how must that limit be related to f(0) ?

 

[HallsofIvy]

You titled this "Find the constant k that will make this piecewise continuous" but then said "Find a value for the constant k that will make this function continous". Those are very different!