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

#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#

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!