Find the limit if it exist?

Question

Find the limit

1) lim 4-sqtroot18-x/x-2

z-->2

2) lim squareroot(7-z) --squareroot(7)/ z

z-->0

Answer 1

lim [ 4 - sqrt(18 - x) ] / [ x - 2 ]

x -> 2

First off, plugging in x = 2 directly would mean the top is equal to zero (since 4 - sqrt(18 - x) = 4 - sqrt(18 - 2) = 4 - sqrt(16) = 4 - 4 = 0, and x - 2 = 2 - 2 = 0).  The form [0/0] is indicative that the limit may exist.

What we want to do is manipulate the expression in a way that it reduces.  

The trick in this case is to rationalize the numerator by multiplying top and bottom by the top's conjugate, 4 + sqrt(18 - x).  This gives us

lim [ 4 - sqrt(18 - x) ] [ 4 + sqrt(18 - x) ] / [ (x - 2)(4 + sqrt(18 - x)) ]

x -> 2

The top becomes a difference of squares, effectively eliminating the radical (square root) symbol.  Don't forget to surround (18 - x) in brackets.

lim [ 16 - (18 - x) ] / [ (x - 2) (4 + sqrt(18 - x) ) ]

x -> 2

Simplify the numerator.

lim [ 16 - 18 + x ] / [ (x - 2)(4 + sqrt(18 - x)) ]

x -> 2

lim [ -2 + x ] / [ (x - 2) (4 + sqrt(18 - x)) ]

x -> 2

Swap the terms in the numerator.

lim [ x - 2 ] / [ (x - 2) ( 4 + sqrt(18 - x)) ]

x -> 2

And now, look at the cancellation we get with (x - 2).

lim [ 1 ] / [ ( 4 + sqrt(18 - x)) ]

x -> 2

And now, we can plug in x = 2 directly.

1 / [ 4 + sqrt(18 - 2) ]

1 / [ 4 + sqrt(16) ]

1 / [ 4 + 4 ]

1/8

Answer 2

Question 1 has been answered, so I'll do 2



multiply numerator and denominator by conjugate

lim [sqrt(7-z) - sqrt(7)]/z

z -> 2

= lim [sqrt(7-z) - sqrt(7)][sqrt(7-z) + sqrt(7)] / z[sqrt(7-z) + sqrt(7)]

z -> 2

= lim [(7-z) - 7] / z[sqrt(7-z) + sqrt(7)]

z -> 2

= lim (-z) / z[sqrt(7-z) + sqrt(7)]  simplify

z -> 2

= lim -1 / [sqrt(7-z) + sqrt(7)]

z -> 2

Now we can plug in z=2

= -1 / [sqrt(7-0) + sqrt(7)]

= -1 / [2sqrt(7)]

Answer 3

About twelve pints and a couple of shorts usually gets to my limit.  I've never tried a lim 4- sqroot - is that an alcopop? Not my kind of booze really.