# I need help to find the value of " m " (variable)

• May 9th 2012, 01:40 PM
reobjan
I need help to find the value of " m " (variable)
In this function y = {sin2x/x (for x no 0), and m-1 (for x=0) } find the value of m that function y to be countinous in x=0.
Can anybody help me with this?
• May 9th 2012, 02:03 PM
jsndacruz
Re: I need help to find the value of " m " (variable)
I'm not sure how advanced your definition of continuity is. In the most elementary terms, a function is continuous at a point b if 1) the limit of the function at b exists and 2) it equals $\displaystyle f(b)$.

You can see that $\displaystyle f(x)= \frac{sin(2x)}{x}$ has a removable discontinuity at $\displaystyle x=0$. So, the limit exists at 0, but the actual value of the function is undefined (because that pesky x in the denominator becomes zero, and dividing by zero is undefined). So, if we simply define $\displaystyle f(0)=2=m-1$, we have defined the function so that the limit at 0 exists, and the function takes on the value of the limits. So, m=3.
• May 9th 2012, 02:03 PM
Plato
Re: I need help to find the value of " m " (variable)
Quote:

Originally Posted by reobjan
In this function y = {sin2x/x (for x no 0), and m-1 (for x=0) } find the value of m that function y to be countinous in x=0.

Hint: $\displaystyle {\lim _{x \to 0}}\frac{{\sin (x)}}{x} = 1$ so $\displaystyle {\lim _{x \to 0}}\frac{{\sin (2x)}}{x} = ~?$
• May 10th 2012, 03:15 PM
HallsofIvy
Re: I need help to find the value of " m " (variable)
Just a little more:
$\displaystyle {\lim _{x \to 0}}\frac{{\sin (x)}}{x} = 1$ so $\displaystyle {\lim _{x \to 0}}2\frac{{\sin (2x)}}{2x} = ~?$[/QUOTE]