Question

another question in calculus regarding infinite limits

Answer 1

questions 1 ad 4 please

Answer 2

@hartnn

Answer 3

@zepdrix

Answer 4

Hey c:

Answer 5

1 and 4 and 7 with detailed explanation please

Answer 6

\[\Large\rm \lim_{x\to\infty}\frac{x^2-2x+3}{x^3+4}\]So we looking at this one first?

Answer 7

yeah

Answer 8

Umm I'm trying to remember how we do these algebraically.
Generally I like to just think of the numerator and denominator separately.

So we have \(\Large\rm x^2-2x+3\) on top,
and \(\Large\rm x^3+4\) on the bottom.

The bottom being a cubic means it will head towards infinity much much faster than the top. So the bottom is "winning".\[\Large\rm \lim_{x\to\infty}\frac{x^2-2x+3}{x^3+4} \qquad\to\qquad \frac{c}{\infty}\]So it's approaching zero, because the bottom is blowing up very quickly.

If we need a more detailed approach, mmm lemme think..

Answer 9

an example is here

Answer 10

Yah that's usually a helpful step, dividing `top and bottom` by the greatest power of x in the denominator.

I'm not sure if it's going to help us much here.
Let's try though.

Dividing everything by x^3 gives us,\[\Large\rm \lim_{x\to\infty}\frac{\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x^3}}{1+\frac{4}{x^3}}\]Yah maybe this approach is alright.
It gives us a bunch of zeros.\[\Large\rm =\frac{0-0+0}{1+0}\]

Answer 11

\[\Large\rm =\frac{0}{1}=0\]

Answer 12

so by evaluation the solution becomes zero ?

Answer 13

Yes, the expression is approaching zero as x approaches infinity.

Answer 14

can i know the difference between both problems,

Answer 15

the example problem had so many steps in it

Answer 16

this is the contunity of the problem

Answer 17

Well in the example problem, they're being very very detailed with the limit rules.
So lemme slow down if I went to quickly.
This was our first step, dividing the top and bottom by the greatest power of x in the denominator,\[\Large\rm \lim_{x\to\infty}\frac{x^2-2x+3}{x^3+4}\color{royalblue}{\left(\frac{\frac{1}{x^3}}{\frac{1}{x^3}}\right)}\]It gave us,\[\Large\rm \lim_{x\to\infty}\frac{\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x^3}}{1+\frac{4}{x^3}}\]If we want to be detailed like the example was, we could split this up and apply our rules.
So for our numerator we have:\[\large\rm \lim_{x\to\infty}\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x^3}=\lim_{x\to\infty}\frac{1}{x}-\lim_{x\to\infty}\frac{2}{x^2}+\lim_{x\to\infty}\frac{3}{x^3}\]\[\large\rm =\lim_{x\to\infty}\frac{1}{x}-2\lim_{x\to\infty}\frac{1}{x^2}+3\lim_{x\to\infty}\frac{1}{x^3}\]\[\Large\rm =0-2\cdot0+3\cdot0\]For the numemator ^

Answer 18

thank you and for the numerator do we need to divide by x^2 or X^3

Answer 19

That was our numerator. We divided everything by x^3.

The denominator, also dividing by x^3 gave us:\[\Large\rm \frac{1}{x^3}(x^3+4)\quad=\quad 1+\frac{4}{x^3}\]

Answer 20

in the numerator the highest power is 2 right . so we need to divide by x^2 ? or x^3

Answer 21

we have to lool at the greatest power of x in both num and den ?

Answer 22

We're dividing both numerator and denominator by the SAME THING.
No no, we don't look at both :)

We divide by the `greatest power of x in the denominator`.

Answer 23

in the example prob they divided by x^2 and also by X?

Answer 24

No they divided by \(\Large\rm \sqrt{\frac{1}{x^2}}\) and \(\Large\rm \frac{1}{x}\). These are of the same degree.
Notice that they didn't divide by \(\Large\rm \frac{1}{x^2}\).

Answer 25

i got yeah !!

Answer 26

thank you

Answer 27

question 17 please

Answer 28

\[\Large\rm as~ x\to\infty,\quad f(x)\to\infty,\quad g(x)\to-\infty\]For part a),\[\Large\rm \lim_{x\to\infty}f(x)+g(x)=\infty\]
Ummmm...
So kind of like what I was talking about before.. in this setup, we would like our f(x) to be "winning". We want f(x) to be approaching `infinity` faster than g(x) is approaching `negative infinity`. That way the sum will still be approaching `infinity`.

Answer 29

So how bout something like this:
\(\Large\rm f(x)=x^2\)
\(\Large\rm g(x)=-x\)

Answer 30

ok lets take that example ..

Answer 31

what do i need to do now

Answer 32

Plugging them in gives us:
\[\Large\rm \lim_{x\to\infty} x^2-x\]Grr I can't remember my algebra for limits :O(
I intuitively know that this is approaching `positive infinity` because of the x^2 term.
When you have a `polynomial`, you only need to look at the largest degree of x to see what the function will do. That is the driving force of what is happening. Everything else becomes insignificant as x gets huge.

Answer 33

Bahh I gotta get to sleep >.<
Maybe miss @myininaya or someone can help you finish these up c:

Answer 34

ha,,,, thanks anyways !!!! thank you so much

Answer 35

@Hero @hartnn

Answer 36

@midhun.madhu1987

Answer 37

@dumbcow

Answer 38

what do you need help with?
do you still need answers to #1 and 4

Answer 39

question 17 and 10 please

Answer 40

ok #10

\[\lim_{x \rightarrow \infty} \sqrt{x^2 +2x} -x\]
\[= \lim_{x \rightarrow \infty} \sqrt{x} \sqrt{x+2} - x\]
\[= \sqrt{x} \sqrt{x} - x \]
\[= 0\]

Answer 41

#17

let f(x) be an increasing function, g(x) be a decreasing function
\[f(x) = x^m\]
\[g(x) = -x^n\]
\[f(x) + g(x) = x^m - x^n\]
If m>n, then limit is positive infinity
If m <n , limit is negative infinity
If m=n, limit = 0

Answer 42

thank you so sos so much sir !!!

Answer 43

question 17 is done ?

Answer 44

do we need a much more detailed explanation for it

Answer 45

yeah i left in a general form, just pick values for m and n

Answer 46

can yeah give me a detailed explanation if possible ?

Answer 47

for a , b and c

Answer 48

ok if the higher exponent is positive then the sum "f+g" will always be positive for large x values
---> 1000^3 - 1000^2 = large positive number

if higher exponent is negative, then sum will always be negative for large x value
--> 1000^2 - 1000^3 = large negative number

if exponents are same then sum equals 0

make sense

Answer 49

no no i get what yu say , but when i need to present it in a paper , how do i write it in a perfect manner

Answer 50

do i just need to give values for m and n ?

Answer 51

thats for you to do, you have to show your own work
i will help with learning the concepts

Answer 52

ha ,,, it will be great if u could do that for me too !!

Answer 53

for parts a b and c , they want you to provide 2 functions f(x) and g(x) which satisfy the given limits

i laid out the general form f(x) and g(x)

Answer 54

ok i got you !!!

Answer 55

this one please

Answer 56

1, 11 , 14 , 15 and 16

Answer 57

SIR?

Answer 58

i dont understand #1 ... cos(infinity) is indeterminate

#11 --> divide everything by n^2

Answer 59

DIVIDE EVERYTHING AND THEN APPLY N TENDS TO INFINITY AND SOLVE IT ?

Answer 60

yes you should get limit = 1

Answer 61

IS IT THE SAME PROCESS WITH THE OTHER QUESTIONS TOO ?

Answer 62

CAN YOU HELP ME WITH 15 AND 16\

Answer 63

@myininaya

Answer 64

please help me with 15 and 16

Answer 65

@myininaya

Answer 66

this screen shot questions 11 14 15 and 16

Answer 67

@iambatman

Answer 68

im a student in 9yh grade talin algebra 1.... I cant help... sorry

Answer 69

ha,,,, i am so wrong,, never saw ur profile!! sorry

Answer 70

its cool

Answer 71

which question do you need help with?

Answer 72

didn't you get help with a question that was similar to 11?

Answer 73

What do you need to divide the top and bottom by in 11?

Answer 74

what term has the high exponent in the bottom?

Answer 75

n^2

Answer 76

so divide top and bottom by n^2

Answer 77

i got it,,, i need the hints for 15 and 16

Answer 78

do you know that -1<=sin(x)<=1?

Answer 79

unaware

Answer 80

what has that relation got to to with the question?

Answer 81

so you didn't know the range of sin(x) is -1 to 1?

Answer 82

doesn't it contain sin(x)?

Answer 83

don't you have \[\lim_{x \rightarrow \infty}(3+\sin(x))x\]

Answer 84

sinx is a y coordinate on the unit circle

Answer 85

x^2 + y^2=1

Answer 86

have you ever notice the value of y on the unit circle stays between -1 and 1?

Answer 87

yes i do

Answer 88

ok so the max value of sin is 1
the min value of sin is -1
then:
\[(3+(-1))x \le (3+\sin(x))x \le (3+1)x\]
I will leave it to you now to use squeeze thm :)

Answer 89

and that inequality is true for really huge numbers

Answer 90

we are only thinking about really huge numbers because of the x->inf