Some Limit Problems?

Question

A] Find the limit

lim　　( 2 - 5 / (x -1)^2 )
x > 1

This is 2 minus 5 over x-1 squared



B] Find the limit

lim　　　　1 - √ (2x^2 ) - 1
x > 1　　　----------------------
　　　　　　　　　　x - 1

The sqrt only contains the 2x^2

For B, this is a type of question way before we learned about L'Hopital.

Answer 1

A) The limit doesn't exist.

B) The limit is 0/0 so use L'hopital's rule

[1-2x/√(2x²-1)]/1 = 1 since 2x/√(2x²-1) -> 0 as x->1

----

OK in B if the square root term is only √(2x²) then the limit does not exist.

If the limit is 0/0 or ∞/∞ then you can use l'hopital's rule (take the derivative of each side) to calculate a limit, but if the result is ∞/k where k is not ∞ or k/0 where k is not 0 then the limit does not exist.

Answer 2

Hello

A. When we use larger values of x (such as 10,1000,100000) then 5/(x-1)^2 approaches zero. The denominator gets larger and larger. Thus the limit would be 2 - 0 = 2.

Also - when you graph the equation or use the table function you will see the limit as x>1 is 2.


B. Lets simplify to get ((2x^2 ))/(x-1) and then simplified more we get -x√ 2 / (x-1). As we use larger values of x -- we see that the top (numerator) will always be larger than the denominator and approaches at a ratio of -1.414.

Also - when you graph the equation or use the table function you will see the limit as x>1 is -1.414.


Hope this helps

Answer 3

1) as (x-1)^2 has lim 0+ -5/(x-1)^2 has limit - infinity and the sum with 2 has limit m- infinity
2)The numerator has limit -sqrt(2) and the denominator has limit 0
so the limit is -infinity from the right and +infinity from the left