How do you determine the limit of #sqrt(x-4)/(3x+5)# as x approaches negative infinity?

Question

Christopher Agresta · Accepted Answer

#1/3#

Notice that you have a square root in your numerator. This can complicate the usual process, so you'll need to take a few extra steps to make sure this works.

The first step is the same as usual: we're going to divide everything by the highest power, which in this case is #x^1#:

#lim_(x->-oo)(1/x)/(1/x) * sqrt(x-4)/(3x + 5)#

For the numerator, we have to make one small simplification to get the right number into the square root:

#=> lim_(x->-oo)(1/sqrt(x^2))/(1/x) * sqrt(x-4)/(3x + 5)#

#=> lim_(x->-oo) sqrt((x-4)/x)/((3x + 5)/x)#

Now, we divide through and simplify:

#=> lim_(x->-oo) sqrt((1 - 4/x))/(3 + 5/x)#

Now, you just go ahead and take your limit:

#=> sqrt((1 - cancel(4/x)))/(3 + cancel(5/x))#

#=>1/3#

Hope that helped :)

Samantha Adkison · Answer

undefined To determine the limit of sqrt(x-4)/(3x+5) as x approaches negative infinity, we can divide both the numerator and denominator by x. This gives us sqrt(1-4/x)/(3+5/x). As x approaches negative infinity, 4/x approaches 0 and 5/x approaches 0. Therefore, the limit simplifies to sqrt(1-0)/(3+0), which is equal to sqrt(1)/3. Thus, the limit of sqrt(x-4)/(3x+5) as x approaches negative infinity is 1/3.