Calculating Limits: Examples and Strategies for Homework Assignments

[mathmathmad]

Homework Statement

1) limit of [ sqrt (1+2x) - sqrt (1-3x) ] / x as x tends to 0
2) limit of (3x^4 -8x^3 + d) / (x^3 - x^2 - x + 1) as x tends to 1
3) limit of [ sqrt (x^2 + 1) - sqrt (x^2 - 1) ] as x tends to infinity

Homework Equations

The Attempt at a Solution

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1) a little confusion here... is the limit 0 or 5/2?
2) since the function is continuous at x0=1, then sub. 1 into all x and the limit does not exist since the denominator = 0 (is it correct?)
3) divide each term inside the sqrt by x^2 and get sqrt 1 - sqrt 1 = 0

 

[Mark44]

Homework Statement

1) limit of [ sqrt (1+2x) - sqrt (1-3x) ] / x as x tends to 0
2) limit of (3x^4 -8x^3 + d) / (x^3 - x^2 - x + 1) as x tends to 1
3) limit of [ sqrt (x^2 + 1) - sqrt (x^2 - 1) ] as x tends to infinity

Homework Equations

The Attempt at a Solution

​

1) a little confusion here... is the limit 0 or 5/2?

Multiply by 1 in the form of [sqrt(1 + 2x) + sqrt(1 - 3x)] over itself.

2) since the function is continuous at x0=1, then sub. 1 into all x and the limit does not exist since the denominator = 0 (is it correct?)

It depends on what d is. Or is that a typo?

3) divide each term inside the sqrt by x^2 and get sqrt 1 - sqrt 1 = 0

No, that won't work, because what you have now is x(sqrt(1 + 1/x^2) - sqrt(1 - 1/x^2)). This is indeterminate as x --> infinity. The thing to do is to multiply by 1 in the form of the conjugate over itself.

 

[mathmathmad]

1) yeah that's what I did and get 5/2 but I thought the answer would be 0
2) yes it is a typo, it should be 5 instead of d. but the limit does not exist right? since the denominator = 0
3) oh! mupltiply by (sqrt (x^2 + 1) + sqrt(x^2-) ) expand and get x^2 + 1 - (x^2 -1) = 0?

 

[Mark44]

1) yeah that's what I did and get 5/2 but I thought the answer would be 0
2) yes it is a typo, it should be 5 instead of d. but the limit does not exist right? since the denominator = 0
3) oh! mupltiply by (sqrt (x^2 + 1) + sqrt(x^2-) ) expand and get x^2 + 1 - (x^2 -1) = 0?

For 1, if you show your work, I'll check it.
For 2, the limit is -oo.
For 3, x^2 + 1 - (x^2 -1) != 0. Check your algebra!

 

[mathmathmad]

1) multiply the equation by (sqrt (1+2x) + sqrt (1-3x)) / (sqrt (1+2x) + sqrt (1-3x))
and get 5x / x (sqrt (1+2x) + sqrt (1-3x))
and get 5 / (sqrt (1+2x) + sqrt (1-3x))
sub. 0 into x and get 5/2

2) how did the limit turn to be negative infinity?

3) oops I'm sorry. I just realized it when I did it on the paper AFTER I posted that.
if I multiply the whole equation by (sqrt (x^2 + 1) + sqrt (x^2-1)) / (sqrt (x^2 + 1) + sqrt (x^2-1) ), get
2 / (sqrt (x^2 + 1) + sqrt (x^2-1) )
then divide each term in the sqrt by x^2?
and get 2/infinity = 0?

does 0 / infinity = 0?
and what is 0 x infinity?

 

[Mark44]

1) multiply the equation by (sqrt (1+2x) + sqrt (1-3x)) / (sqrt (1+2x) + sqrt (1-3x))
and get 5x / x (sqrt (1+2x) + sqrt (1-3x))
and get 5 / (sqrt (1+2x) + sqrt (1-3x))
sub. 0 into x and get 5/2

Right, the limit is 5/2. On a technical note, you're not multiplying an equation by that stuff; you're multiplying an expression by that stuff. An equation always has an = sign lurking about.

2) how did the limit turn to be negative infinity?

Actually, my answer was incorrect. The limit doesn't exist at all. The left-side limit is oo and the right-side limit is -oo. To see what's happening around x = 1, notice that you can factor (x - 1) out of both the numerator and denominator.

3) oops I'm sorry. I just realized it when I did it on the paper AFTER I posted that.
if I multiply the whole equation by (sqrt (x^2 + 1) + sqrt (x^2-1)) / (sqrt (x^2 + 1) + sqrt (x^2-1) ), get
2 / (sqrt (x^2 + 1) + sqrt (x^2-1) )
then divide each term in the sqrt by x^2?
and get 2/infinity = 0?

does 0 / infinity = 0?
and what is 0 x infinity?

Prob 3 is pretty clear-cut. After multiplying the expression by the conjugate over itself, you get 2/(sqrt(x^2 + 1) + sqrt(x^2 - 1)). As x --> oo, the denominator gets large without bound, but the numerator is stuck at 2, so the whole fraction approaches 0.