If a1=1 an+1=[4+3an] ÷[3+2an]
n>=1. And an>0 if limit n infinity an =a
Then a =?
Answers
Answer:1. In the special theory of relativity, the mass of a particle with velocity v is given by
m = p
m0
1 − v
2/c2
where m0 is the mass of the particle at rest and c is the speed of light. What happens as
v → c
−?
Answer: As v → c
−, the fraction
v
2
c
2
gets closer and closer to 1 (though it is always less than 1 since v is approaching c from the
left). Hence, the quantity
1 −
v
2
c
2
is approaching zero. Therefore, as v → c
−, the mass:
m = p
m0
1 − v
2/c2
→ +∞.
2. Evaluate
lim
t→0
1
t
−
1
t
2 + t
.
Answer: Notice that t
2 + t = t(t + 1). Therefore,
1
t
−
1
t
2 + t
=
t + 1
t(t + 1) −
1
t(t + 1) =
t
t(t + 1) =
1
t + 1
.
Therefore,
lim
t→0
1
t
−
1
t
2 + t
= lim
t→0
1
t + 1
= 1.
3. Evaluate
limx→∞
x
√
x
2 + 1
.
Answer: Using L’Hˆopital’s Rule,
limx→∞
x
√
x
2 + 1
= limx→∞
1
1
2
√
1
x2+1
· 2x
= limx→∞
1
√ x
x2+1
= limx→∞
√
x
2 + 1
x
,
provided that limit exists. If we knew the limit existed, then it would have to be 1, since the
left and right sides are reciprocals of each other. However, we don’t know, a priori, that the
limit exists, so we must use another method.
Let’s divide both the numerator and denominator by x:
Step-by-step explanation: