Propose a method for finding the limit of h(x) = (2x^3 - 5x + 3)/(x^2 - 1) as x approaches 1.
user
Sure, one way to find this limit is to apply L'Hôpital's rule, which states that if you have an indeterminate form such as infinity/infinity when x approaches a, you can find the limit by taking the derivative of the numerator and denominator and evaluating them at x = a.

Applying this rule, we can simplify the expression as follows:

h'(x) = (6x^2 - 5)(2x - 1)/(x^2 - 1)^2

Now, as x approaches 1, the numerator becomes indeterminate due to the infinity/infinity form. However, we can use L'Hôpital's rule again on numerator, resulting in:

h''(x) = (27x - 75)(2x - 1)/(x^2 - 1)^3

In this way, the limit we're trying to find can be obtained by evaluating the expression as x goes to 1:

lim x->1 h''(x) = (27*1 - 75)(2*1 - 1) / (1^2 - 1)^3 = -4*24 / 0 = -96

So, the limit of h(x) as x approaches 1 is -96. 

Alternatively, we can divide both the numerator and denominator by the factor (x^2 -1), which simplifies the expression and makes it an indeterminate form as before:

h(x) = ((2x^3 - 5x + 3) / (x^2 -1)) * (x^2 -1) / (x^2 -1)

= (2x^3 - 5x + 3) / (x - 1)(x+1)

Using L'Hôpital's rule combined with separating the linear term from the cube factor we have:

h'(x) = 6x² - 5 or \[h'(x) = (6(3x - 1)(2x +1) / (x - 1)(x + 1)) = (27x^2 - 75)(2x - 1)(2x + 1) / ((x - 1)(x + 1))^3)

Using this simplification, we also get the same aforementioned result as for the first derivative. Note that this last method is more complex due to the additional step but it demonstrates an answer without the need to repeat the L'Hôpital's trick for finding the end result, only computation needed is for defining the original indeterminate and afterwards the own formula if it represents a fraction making the L' hospice extraordinary simple.
assistant
Propose a method for finding the limit of h(x) = (2x^3 - 5x + 3)/(x^2 - 1) as x approaches 1.
user
To find the limit of h(x) = (2x^3 - 5x + 3)/(x^2 - 1) as x approaches 1, we first try to simplify the expression by factoring both the numerator and denominator.

For the numerator, we can use synthetic division or the Remainder Theorem to find that (x - 1) is a factor since h(1) = 0. Then, perform polynomial long division or synthetic division to get:

2x^3 - 5x + 3 = (x - 1)(2x^2 + 2x - 3)

Now, let's factor the denominator:

x^2 - 1 = (x + 1)(x - 1)

So now we have:

h(x) = [(x - 1)(2x^2 + 2x - 3)] / [(x + 1)(x - 1)]

Since x - 1 appears in both the numerator and denominator, we can cancel it out:

h(x) = (2x^2 + 2x - 3) / (x + 1)

Now, we can find the limit as x approaches 1 without any issues:

lim (x -> 1) h(x) = lim (x -> 1) [(2x^2 + 2x - 3) / (x + 1)]

Plug in x = 1:

= (2(1)^2 + 2(1) - 3) / (1 + 1)
= (2 + 2 - 3) / 2
= 1

So, the limit of h(x) as x approaches 1 is 1.
assistant