How do you find $lim (y+1)/((y-2)(y-3))$ as $x->3^+$ ?

Question

1682 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-06T22:12:00

$lim_(y to 3^+) f(y) =oo$

let $f(y) = (y+1)/( (y-2)(y-3) )$

we know as $y to 3^+ , y + 1 to 4$
we know as $y to 3^+ , y -2 to 1$
we know as $y to 3^+ , y - 3 to 0$

for this problem we are going to write $f(y)$ as;
$(y+1) /( y -2) 1/(y-3) = f(y)$

we know that $(y+1)/(y-2) to 4$ as $y to 3^+$

But now we must consider $1/(y-3)$

as $y to 3^+$ the demominator gets very small $( to 0 )$ but as $y to 3^+$ it is possotive, so hence as denominator $to 0$ , $1/(y-3) to oo$

So hecne $4 * oo = oo$

= $oo$

How do you find lim (y+1)/((y-2)(y-3))lim (y+1)/((y-2)(y-3)) as x->3^+x->3^+?