anybody up for some calculus?

Question

i have 2 problems and i would really appreciate it if anyone could help me

1) as x goes to 1, find the limit of (x^3 -1) / (x^4 -1)

2) find f(g(x))
f(x)= x^2 +1 and g(x)=1/(x^2 +1)

Answer 1

Solution:1 (X^3-1)/((x^4-1) equals

((x-1)(x^2+X+1))/(X^2+1)(X+1)(x-1)

Cancelling out x-1 from the numerator and denominator,you get (x^2+X+1)/((X^2+1)(X+1)).

Now substituting x=1,you get 3/4 or 0.75

Solution:2
f[g(x)]=f[1/(1+x^2)]=(1/(x^2+1))^2 + 1...and simplify it to get the result by taking LCM.

Answer 2

The numerator = (x-1)(x^2+x+1)
the denominator = (x-1)/(x^3+x^2+x+1)
so the reuslt = (x^2+x+1)/(x^3+x^2+x+1)

as x-> 1 the value is (1+1+1)/(1+1+1+1) = 3/4
problem 2
f(x) = x^2+1
g(x) = 1/(x^2+1)

g(f(x)) = 1/(f(x)^2+1)
= 1/((x^2 +1)^2 +1)
and if you want to simplify
=1/(x^4+2x^2 + 1 +1) = 1/(x^4+2x^2+2)

Answer 3

The first problem requires using l'Hospital's rule. Differentiating top and bottom with respect to x, we get 3x^2/4x^3, which at x = 1 is 3/4. It will be instructive to evaluate (using your pocket calculator) the value of the expression at x = 1.001. Previous responders get the correct numerical value, but by an invalid procedure: multiplying or dividing by 0/0 (which is what they are doing) is not defined.
Second problem requires no calculus, it is just a matter of substitution. f(g(x)) = (1/(x^2 +1))^2 + 1.

Answer 4

(1) To find lim(x^3 -1) / (x^4 -1) as x goes to 1, use L'Hopital's Rule.
By doing that we get lim(3x^2)/(4x^3)
As x goes to 1, the second limit (which equals the first limit) tends to (3*1^2)/(4*1^3) = 3/4

(2) f(g(x)) = f(1/(x^2 +1)) = 1/(x^2 +1)^2 + 1

Answer 5

Problem no. (1)....

As x approaches to 1,
lim (x^3 - 1) / (x^4 - 1)
=lim (x^3 - 1) / {(x^2+1)(x+1)(x-1)}
=lim [1 / {(x^2+1)(x+1)}] lim [(x^3 - 1) / (x-1)]
= [1 / {(1^2+1)(1+1)}] lim [(x^3 - 1^3) / (x-1)]
= [1 / {2*2}] 3*1^(3-1) [according to limit properties]
= [1 / 4]*3*1^2
= 3 / 4

Problem no. (2).....

f(g(x))
=f( 1 / (x^2+1) ) [because, g(x) = 1 / (x^2+1)]
=1 / (x^2+1)^2 + 1 [because, f(x) = x^2+1]
={1+(x^2+1)^2}/(x^2+1)^2
={x^4+2x^2+2} / {x^4+2x^2+1}

Answer 6

1. Substitute 1 to the equation and you will get 0/0 which is infinity. However, the logical answer to your limit shall be 1 because x is equals to a definite value

2. Substitute the g(x) inside the f(x) then you will get
(1/(x^2+1))^2 + 1 for your answer

Answer 7

1. Since both the numerator and denominator are 0 when you plug in 1. (x-1) is a factor of both. factor out an (x-1) from both and go from there.

2. in the equation f(x) replace x with g(x).

Answer 8

you will desire to be attentive to that the spinoff of a relentless function is 0 and that of exponential function e^x is itself. nice humorous tale with a twist on the top. Now, who choose to enlarge it?