CALCULUS about LIMITS?

Question

Evaluate the limit


4) lim square root of x-2 over square root of x^2 - 4x -4
X-->2
5) lim sq root of 1- 8x^3 over sq rt 1-4x^2
X--->1/2
6) lim sq rt of x -sq rt of 2 over x^2 - 2x
x--->2
7) lim 2 - cube root of x over (sq rt of x - 4) - 2
x---->8
8) lim x^2/3 -(2sq rt of x) over 4 - (16/x)
x--->8

Answer 1

Hi,

I gave 4,5, and 6 a shot but I doubt 6 is right. I hope you get a good clear answer. I'd like to see it.

4) lim square root of x-2 over square root of x^2 - 4x -4
X-->2

lim = (2 - 2)/(2² -4(2) - 4) = 0/-8 = 0

5) lim sq root of 1- 8x^3 over sq rt 1-4x^2
X--->1/2

1-8x³....(1-2x)(1+2x+4x²)
-------.=.----------------------- =
1-4x²....(1-2x)(1+2x)

........(1+2x+4x²)....1 +2(½)+4(½)²
lim----------------- =.-------------------- =
x->½...(1+2x)........1 +2(½)

1 +2(½)+4(½)²....1+1+1
------------------- =.---------- = 3/2 <== answer
..1 +2(½)..............1 + 1


6) lim sq rt of x -sq rt of 2 over x^2 - 2x
x--->2
.._______
√{x - √(2)}
--------------- =
x² - 2x

.._______
√{x - √(2)}
---------------
x(x - 2)

.._______
√{x - √(2)}
------------------------- =
x{x - √(2)}{x + √(2)}

{x - √(2)}^½
------------------------- =
x{x - √(2)}{x + √(2)}

..1
--------------------------------- =
x{x - √(2)}^(½)*{x + √(2)}

.............1
lim....--------------------------------- =
x->2...x{x - √(2)}^(½)*{x + √(2)}

.............1
lim....--------------------------------- =
x->2...2{2 - √(2)}^(½)*{2 + √(2)}

......1
---------------------
...4{2 + √(2)}^½

7) lim 2 - cube root of x over (sq rt of x - 4) - 2
x---->8
?
8) lim x^2/3 -(2sq rt of x) over 4 - (16/x)
x--->8

8^(2/3) -2√8
----------------- =
4 - 16/8

4 -4√2
--------- =
4 - 2

4 -4√2
--------- =
....2

2 -2√2



I hope that helped!!

Answer 2

These are really just algebra questions. In each one, you need to factor & reduce things to see whether you are really "blowing up" at the limit value.

E.g., 4) √(x-2) / √(x^2 - 4x - 4) = √(x-2) / √(x-2)^2 = √(x-2) / (x - 2) = 1 / √(x - 2). So as x approaches 2 this really is dividing by 0, thus the limit does not exist.

5) √(1 - 8x^3) / √(1 - 4x^2) = √((1 - 2x)(1+2x + 4x^2)) / √((1 - 2x)(1 + 2x))
= √(1+2x+4x^2) / √(1 + 2x). This can be evaluated at 1/2, giving a limit of √(3/2).

Added later: The third responder seems to have overlooked the square roots on 4 & 5. The answer for 8 is correct. Here's one more.

6) The denominator x^2 - 2x factors as x(x - 2). The trick is to consider x - 2 as a difference of squares, giving x(√x - √2)(√x + √2). Then the ratio simplifies as (√x - √2) / (x(√x - √2)(√x + √2)) = 1 / (x(√x + √2) which gives a limit value of 1 / (4√2).

Answer 3

I'll work this one.

8) lim x→8 of [x^(2/3) - 2√x] / [4 - 16/x]
= [8^(2/3) - 2√8] / [4 - 16/8] = (4 - 4√2) / (4 - 2)
= 4(1 - √2) / 2 = 2(1 - √2)

Answer 4

5.♠ thus √((1-8x^3)/ (1-4x^2)) =
= √((1+2x +4x^2)/(1+2x))= √1.5;
the rest are easy, do them your self!