Question #4186

Let (x sub n) := & 1/ln(n+1) for n element of natural numbers.&

1. use the definition of limit to show that lim(x sub n)=0&

2. find the specific value of k(e) as required in the definition of limit for each of & (i) e = 1/2 and & & (ii) e=1/10.

1. use the definition of limit to show that lim(x sub n)=0&

2. find the specific value of k(e) as required in the definition of limit for each of & (i) e = 1/2 and & & (ii) e=1/10.

Expert's answer

Task.Let (x sub n) := 1/ln(n+1) for n element of natural numbers.

1. use the definition of limit to show that lim(x sub n)=0

Proof.

We have to show that for any e>0 there exists and integer k(e)0

such that for all n > k(e) we have that

|x_n| < e

that is

|1/ln(n+1)| < e,

and so

(*) ln(n+1) > 1/e.

Since

lim_{x->+infinity} ln(x+1) = + infinity,

for any A>0 there exist an integer number M(A)>0 such that

(**) ln(m+1) > A for all m>M(A)

It remains to put A=1/e and k(e)=M(A).

Then (*) holds true.

=================================

2. find the specific value of k(e) as required in the definition of limit for each of

(i) e = 1/2 and

(ii) e=1/10.

Solution.

We should find such k(e) that

ln(m+1) > 1/e

for all m>k(e), and so

m > exp(1/e)-1

Thus we can put

k(e) = [exp(1/e)],

where [t] means the integer part of t.

Now if e=1/2, then

k(1/2) = [exp(2)] = [7.38905609893065] = 7.

Now if e=1/10, then

k(1/10) = [exp(10)] = [22026.4657948067] = 22026

1. use the definition of limit to show that lim(x sub n)=0

Proof.

We have to show that for any e>0 there exists and integer k(e)0

such that for all n > k(e) we have that

|x_n| < e

that is

|1/ln(n+1)| < e,

and so

(*) ln(n+1) > 1/e.

Since

lim_{x->+infinity} ln(x+1) = + infinity,

for any A>0 there exist an integer number M(A)>0 such that

(**) ln(m+1) > A for all m>M(A)

It remains to put A=1/e and k(e)=M(A).

Then (*) holds true.

=================================

2. find the specific value of k(e) as required in the definition of limit for each of

(i) e = 1/2 and

(ii) e=1/10.

Solution.

We should find such k(e) that

ln(m+1) > 1/e

for all m>k(e), and so

m > exp(1/e)-1

Thus we can put

k(e) = [exp(1/e)],

where [t] means the integer part of t.

Now if e=1/2, then

k(1/2) = [exp(2)] = [7.38905609893065] = 7.

Now if e=1/10, then

k(1/10) = [exp(10)] = [22026.4657948067] = 22026

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