Evaluate the limit, if it exists

Question

lim (x --> 16)  (4 - √x)/(16x - x²)

 

 

p107, #27

ANSWER: - 1/128

Answer 1

Hi Mia,

 

I am guessing you might be at a place in your Calculus course where you are not yet aware of L'Hospital's rule.

 

Another way to solve this problem is to change the way the expression looks before substituting 16 for x. Try multiplying numerator and denominator of the original fraction by the conjugate of the numerator (4 + √x).

 

That will give in the numerator (16-x)  (difference of squares).

 In the denominator factor out the x and also multiply by the conjugate giving x(16-x)(4+√x).

 

So we have (16-x)/[x(16-x)(4+√x)]. This reduces to 1/[x(4+√x)]. Find the limit by substituting 16 for x.

 

lim_x->16 1/[x(4+√x)] = 1/[16(4+4)] = 1/128.

Answer 2

Since plugging in 16 for x give us indeterminate form we can use L'Hopitals Rule that says the derivative of the numerator over the derivative of the denominator will be equal to our limit as x approaches some value. 

 

d/dx[4-sqrt(x)]=-(1)/(2sqrt(x)) <---numerator

d/dx[16x-x²]=16-2x<----denominator

 

Then plug in 16 for x and we get

 

-[(1/(2sqrt(16))/(16-2(16))=-(1/8)/(-16)

=1/128

Answer 3

lim            (4 - x^1/2) / (16x - x²)

x-->16

 

 

 

Use L'Hospital Rule.

 

lim            (-1/2)x^-1/2 / (16 - 2x)

x-->16

 

 

Now we can plug in x=16 to find the limit.

 

lim           (-1 / (2 * √16)) / (16 - 2(16))

x-->16

 

 

lim           (-1 / 8) / (16 - 32)

x-->16

 

 

lim            (-1 / 8) / (-16)

x-->16

 

 

lim             (-1 / 8) * (-1 / 16)   

x-->16

 

 

lim            1 / 128

x-->16