Limits Involving Infinity
Key Questions

Answer:
#lim_(x>oo)sqrt(x) = oo# Explanation:
Intuitively, as there is no bound to how large we can make
#sqrt(x)# by increasing#x# , we expect that the limit as#x>oo# of#sqrt(x)# would be#oo# . Indeed, if there were such a bound, say#x_0# , then we would arrive at a contradiction, as#sqrt(x_0^2+1) > sqrt(x_0^2)= x_0# .We can, however, approach the problem in a more rigorous manner.
We say that the limit as
#x>oo# of a function#f(x)# is#oo# (alternately#f(x)>oo# as#x>oo# ), denoted#lim_(x>oo)f(x)=oo# , if, for every integer#N>0# there exists an integer#M>0# such that#x>M# implies#f(x)>N# .Less formally, that means that for any real value,
#f(x)# will be greater than that value for large enough#x# .Our claim is that
#lim_(x>oo)sqrt(x) = oo# . Let's prove it using the above definition.
Take any integer
#N>0# , and let#M=N^2# . Then, for any#x>M# , we have#sqrt(x) >sqrt(M) = sqrt(N^2) = N# We have shown that for any integer
#N>0# there exists an integer#M>0# such that#x>M# implies#sqrt(x) > N# , thereby proving that#lim_(x>oo)sqrt(x) = oo# .
The above method actually can be used to show that
#x^k>oo# as#x>oo# for any#k>0# . If we start with an arbitrary#N>0# and let#M=N^(1/k)# , then for#x>M# we have#x^k > M^k = (N^(1/k))^k=N# . As#sqrt(x) = x^(1/2)# , the above is just a special case of this. 
Answer:
#Lim_(x>oo)x=oo# Explanation:
Break the problem down into words: "What happens to a function,
#x,# as we continue increasing#x# without bound?"#x# would also increase without bound, or go to#oo.# Graphically, this tells us that as we continue heading right on the
#x# axis (increasing values of#x,# going to#oo)# our function, which is just a line in this case, keeps heading upwards (increasing) with no restrictions.graph{y=x [10, 10, 5, 5]}

The answer is undefined.
#lim_(x>oo)sqrt(x)=# undefinedThe reason for this is that the domain of the square root function is
#x>=0# . The limit is undefined if the limit is not being evaluated in the domain. 
Answer:
I usually do this informally.
Explanation:
I ask myself what kinds of numbers do I get if I put in more and more negative numbers for
#x# . (Often described by saying "bigger and bigger negative numbers".) ("Big" means "far from zero".)Examples:
Example 1
#f(x) = 3x^47x^3+2x+72# For very very big numbers, the only term that matters is the largest power term:
#3x^4# . As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?#x^4# will always be positive and when I multiply by#3# the answer will still be positive, so I get bigger and bigger positive numbers.#lim_(xrarroo)f(x) = oo# Example 2
#g(x) = 5x^7+43x^4+2x^35x+21# For very very big numbers, the only term that matters is the largest power term:
#5x^7# . As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?#x^7# will be negative for negative#x# 's and when I multiply by#5# the answer will still be negative, so I get bigger and bigger negative numbers.#lim_(xrarroo)g(x) = oo# Example 3 (last)
#h(x) = 8x^6+7x3# For very very big numbers, the only term that matters is the largest power term:
#8x^6# . As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?#x^6# will be positive for all#x# 's and when I multiply by#8# the answer will become negative, so I get bigger and bigger negative numbers.#lim_(xrarroo)h(x) = oo#