1. ## Limit help

Any hints to how to solve this type of problem would be great

f(x) = 6 + ( x^7 / sin(x) ) for x > 0
f(x) = a for x = 0
f(x) = 1 / ( x+ 9 ) + b for x < 0

(peicewise function)

Find lim x -> 0+ f(x)

find lim x -> 0- f(x)

then how would i find what a and b are?

thanks

2. Originally Posted by Stosw
Any hints to how to solve this type of problem would be great

f(x) = 6 + ( x^7 / sin(x) ) for x > 0
f(x) = a for x = 0
f(x) = 1 / ( x+ 9 ) + b for x < 0
(peicewise function)
Find lim x -> 0+ f(x)
find lim x -> 0- f(x)
then how would i find what a and b are?
thanks
For the limit as x goes to 0 from the positive side you use the first branch of the function, since its domain is x > 0.

To evaluate the limit, write $f(x) = 6 + x^6 \frac{x}{sin(x)}$

and note that $\frac{x}{sin(x)}$ OR $\frac{sin(x)}{x}$ both go to 1 as x approaches 0.

So the limit becomes 6 + 0* 1 = 6

For the other limit use the third branch, which gives $\frac{1}{9} + b$

You haven't shared the conditions this problem wants. I assume it's continuity, which means the limits from the left and right are equal to the function value.

So, two conditions:

$6 = \frac{1}{9} + b$ (limit exists)

$f(0) = a = 6$ (limit equal to function value, so the graph, when you draw it, is filled)

Good luck!!