Find limits algebraically and by manual graphing

Question

lim      x²+6x+5/x²-3x-4

x->-1 

 

lim    (√2x+3)-3/9-x²

x-->3

 

 

Answer 1

For the first limit, you need to parenthesize each quadratic (one in numerator, and one in denominator).

 

Factor them; each has a factor (x+1) which cancels, so you want limit (as x approaches -1) of (x+5)/(x-4).

This can be done by substitution. Answer: -4/5.

 

Similarly, parenthesization is required to ensure the correct PEMDAS interpretation of the second limit. Please clean that up.

 

Answer 2

We can't do graphing here, but we can show how to find limits analytically

lim_x→-1 (x2+6x+5)/(x2-3x-4)=

lim_x→-1[(x+1)(x+5)]/(x+1)(x-4)]=lim_x→-1(x+5)/(x-4)

=(-1+5)/(-1-4)=-4/5

 

lim_x→3(√(2x+3)-3)/(9-x²)=

=lim_x→3(2x-6)/[(3-x)(3+x)(√(2x+3)+3)]=

=lim_x→3-2/[(3+x)(√(2x+3)+3)]=

=-2/[6(3+3)]=-1/18