AP CALCULUS BC Group B

Steps: 

1. Find the closest known of your function by looking at the highest exponents and simplifying

2. Determine if your known diverges or converges for the given interval

3. Find "Big L" by taking the lim (f(x) / g(x)) where f(x) is your unknown and g(x) is your known.

 x->∞                                                           

4. If 0<L<∞ then f(x) acts the same as g(x)

5. Write a sentence to state your conclusion based upon the limit comparison test

                                                  ∞

Example 1: f(x)= ∑ ((n+1)/(1-n+n^2))

                      ∞                     n=1

known: g(x) = ∑  (1/n) which diverges by p series test

                      n=1

L = lim (f(x)/g(x)) = 1 

      n->∞

0<1<∞ so f(x) also diverges                        ∞                                                       ∞

By the LCT, L= lim(f(x)/g(x)) where g(x) is  ∑ (1/n)  which diverges and f(x) is ∑ ((n+1)/(1-n+n^2)). Since 0<1<∞, f(x) diverges because it acts

                        n->∞                                   n=1                                                    n=1

the same as g(x). 

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                                                ∞

Example 2: f(x) = ∑ ((n+1)/(n2^n))

                     ∞                       n=1

known: g(x)= ∑ (1/2^n) which converges by geometric test ( r= 1/2)

                     n=1

L= lim (f(x)/g(x))=1

     n->∞

0<1<∞ so f(x) also converges                     ∞                                                              ∞

By the LCT, L= lim (f(x)/g(x)) where g(x) is ∑ (1/2^n) which converges and f(x) is ∑ ((n+1)/(n2^n)). Since 0<1<∞, f(x) also converges because

                        n->∞                                    n=1                                                          n=1

it acts the same as g(x). 

                               ∞

Example 3: f(x) =  ∑ ((5n^2-n+1)/(n^3 +4n+3)) 

                      ∞                         n=1

known: g(x) = ∑ (1/n) which diverges by p series test

                      n=1

L = lim ((f(x)/g(x)) = 5

     n->∞

0<5<∞ so f(x) also diverges                        ∞                                                      ∞

By the LCT, L= lim (f(x)/g(x)) where g(x) is ∑ (1/n) which diverges and f(x) is ∑ ((5n^2-n+1)/(n^3+4n+3)). Since 0<5<∞, f(x) also diverges

                        n->∞                                    n=1                                                   n=1

because it acts the same as g(x). 

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