Can someone please help me with these two? im so lost and i really appreciate it if someone could show me what to do!
1) To evaluate
lim x→infinity sqrt (x^2 + 4)
first consider that as x becomes infinitely large, x^2+4 -> _____
2)Find lim x→ infinity f(x)if,
for all x > 1,
(6e^x − 17)/3e^x < f(x) < 2sqrt(x)/ sqrt (x − 1)
forget the first one i got it. just the second one i don't know where to begin
for (6e^x - 17)/(3e^x)
divide top and bottom by e^x
we get
(6-17/e^x)/3 , as x ----> ∞ , the expression ---> 2
and 2√x/√(x-1) ----> 2 , as x ---->∞
so as x---->∞
2 < f(x) < 2
which implies f(x) = 2
SOOOO HAPPY YOU ANSWERED ME!
ok now im just trying to process this in my head. i understand dividing it by e^x but you lost me at the expression -->2
sorry to be a pain, i just want to make sure i fully understand this. thank you so much! you are a life saver!
in (6-17/e^x)/3 , as x ----> ∞
what happens when you divide 17 by a very very very large number ?
Doesn't it go to zero ?
so you have (6-0)/3 = 2
for the right side 2√x/√(x-1)
when x is really really large, do you think it makes any difference if we take √x or √(x-1) ?
e.g. take
√123456789 and √123456788
now divide them to get 1.000000004
so the right side would be 2.00000008 or just plain 2
so the f(x) is sandwiched between 2 and 2 , thus it MUST BE 2
Sure! I'll be happy to help you with these two problems. Let's start with the first one:
1) To evaluate the limit lim x→infinity sqrt (x^2 + 4), we need to consider what happens to the expression inside the square root as x becomes infinitely large.
As x approaches infinity, the term x^2 will dominate the expression x^2 + 4. This is because as x becomes very large, the value of 4 becomes insignificant compared to x^2. So, we can say that x^2 + 4 approaches x^2 as x approaches infinity.
Therefore, the limit can be simplified as lim x→infinity sqrt (x^2 + 4) = lim x→infinity sqrt (x^2) = lim x→infinity x.
So, the limit as x approaches infinity of sqrt (x^2 + 4) is infinity.
Now, let's move on to the second problem:
2) The inequality given is (6e^x − 17)/3e^x < f(x) < 2sqrt(x)/ sqrt (x − 1) for all x > 1.
To find the limit lim x→ infinity f(x), we need to determine what happens to both the upper and lower bounds of the inequality as x becomes infinitely large.
Let's start with the lower bound: (6e^x − 17)/3e^x. As x approaches infinity, the exponential term e^x grows much faster than any other term in the expression. So, we can simplify the lower bound as (6e^x)/3e^x = 2.
Therefore, the lower bound approaches 2 as x approaches infinity.
Now, let's consider the upper bound: 2sqrt(x)/ sqrt (x − 1). As x approaches infinity, the term x − 1 becomes insignificant compared to x, and we can ignore it. So, we can simplify the upper bound as 2sqrt(x)/ sqrt (x) = 2.
Therefore, the upper bound also approaches 2 as x approaches infinity.
Since the lower and upper bounds of f(x) both approach 2 as x approaches infinity, we can conclude that the limit lim x→ infinity f(x) is also equal to 2.
I hope this explanation helps you understand how to evaluate these limits and inequalities.