Question

How do you evaluate `sin(x-3)/(x^2+4x-21)` as x approaches 3?

Answer 1

`1/10`

Explanation:

`lim_(xrarr3) Sin(x-3)/(x^2+4x-21)`

Using L-Hospital rule,i.e differentiate numerator and denominator separately without using the quotient rule,
we get,

`lim_(xrarr3) Cos(x-3)/(2x+4) *d/dx(x-3)`

`lim_(xrarr3) Cos(x-3)/(2x+4)*(1-0)`

Putting x=3,

`Cos(3-3)/{2(3)+4`

=`Cos(0)/10`
=`1/10`

Answer 2

The limit is equal to `1/10`.

Explanation:

To start, let's just try plugging `x=3` into the equation and see what happens:

`color(white)=>sin(x-3)/(x^2+4x-21)`

`=>sin(3-3)/(3^2+4(3)-21)`

`=sin0/(9+12-21)`

`=0/0`

The indeterminate `0/0`. In these types of cases, we can apply l'Hopital's rule:

If `lim_(x->a) f(x)/g(x)=0/0` or `lim_(x->a) f(x)/g(x)=(+- infty)/(+- infty)`, then

`lim_(x->a) f(x)/g(x)=lim_(x->a) (f'(x))/(g'(x))`

In other words, take the derivative of the top and bottom of the fraction, then plug in the value. Let's do that:

`color(white)=lim_(x->3) sin(x-3)/(x^2+4x-21)`

`=lim_(x->3) (d/dx(sin(x-3)))/(d/dx(x^2+4x-21))`

`=lim_(x->3) cos(x-3)/(d/dx(x^2+4x-21))`

`=lim_(x->3) cos(x-3)/(d/dx(x^2)+d/dx(4x)-d/dx(21))`

`=lim_(x->3) cos(x-3)/(2x+4-0)`

`=lim_(x->3) cos(x-3)/(2x+4)`

`=cos(3-3)/(2(3)+4)`

`=cos(0)/(2(3)+4)`

`=cos(0)/(6+4)`

`=cos(0)/10`

`=1/10`

That's the limit. We can observe this from the graph of the function:

graph{sin(x-3)/(x^2+4x-21) [-1, 7, -0.02, 0.2]}

Hope this helped!