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Evaluating the Problems using Limits and their Properties.
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Evaluate
lim xโ 4
x^2 x^2 โ 4
If we try direct substitution, we end up with โ^160 โ (i.e., a non-zero constant over zero), so weโll get either +โ or โโ as we approach 4. We then need to check left- and right-hand limits to see which one it is, and to make sure the limits are equal from both sides.
x^2 (x โ 4)(x + 4)
As x โ 4 โ, the function is negative since (+)
2 (โ)(+) = (โ), so the left-hand limit is^ โโ.
x^2 (x โ 4)(x + 4)
As x โ 4 +, the function is positive since (+)
2 (+)(+) = (+), so the right-hand limit is +โ.
Since the left- and right-hand limits are not equal,
lim xโ 4
x^2 x^2 โ 4
Evaluate
lim xโโ 3
x^2 โ 2 x โ 3 x^2 + 6x + 9
If we try direct substitution, we end up with โ^120 โ, so weโll get either +โ or โโ as we approach -3. As in the last example, we need to check left- and right-hand limits to see which one it is, and to make sure the limits are equal from both sides.
(x โ 3)(x + 1) (x + 3)^2
As x โ โ 3 โ, the function is positive since (โ(โ)()โ 2 )= (+)(+) = (+), so the left-hand limit is +โ.
(x โ 3)(x + 1) (x + 3)^2
As x โ โ 3 +, the function is positive since (โ(+))(โ 2 )= (+)(+) = (+), so the right-hand limit is also +โ.
Since the left- and right-hand limits are the same,
lim xโ 4
x^2 โ 2 x โ 3 x^2 + 6x + 9
Evaluate
lim xโ 0 +
sin(x)
First of all, we note that direct substitution fails (we get โ^20 โ). There are a couple of different ways we can look at this problem. For either one, we observe that as x โ 0 +, sin(x) also
goes to zero from values greater than zero (i.e., sin(x) โ 0 +): So, lim xโ 0 +
sin(x)
is either +โ
or โโ. From what we observed above, we know the function will be (+)(+) = (+), so the limit is +โ.
The other way we can approach this is to replace sin(x) with another variable that goes to the same value as sin(x) when we take the limit. Since sin(x) โ 0 +^ as x โ 0 +, then
lim xโ 0 +
sin(x)
= lim tโ 0 +
t
( which still = โ).
and then dividing through by x gives us
x โ 1 x
x โ cos(x) x
x + 1 x
(since x โ +โ, x is positive, so dividing through by x wonโt change the inequalities). Now we can use the Squeeze Theorem to say that
lim xโโ
x โ 1 x
6 lim xโโ
x โ cos(x) x
6 lim xโโ
x + 1 x
Both outside limits involve rational functions with the same degree in both numerator and denominator, so the limit as x โ โ is simply the ratio of the leading coefficients, which in both of these is 11 = 1. Since the outside limits go to the same value, then, by the Squeeze Theorem, lim xโโ
x โ cos(x) x
Evaluate
lim xโโโ
5 x^2 x + 3
Note: In this case we canโt use the theorem we talked about in class for the limit of a rational function since that theorem only applied in cases where x โ +โ, not when x โ โโ. However, we can still use the method of dividing through by a power of x. Now, we donโt always want to divide through by the highest power from either numerator or denominator (in this case, if we divided the numerator and denominator through by x^2 , weโd end up with a numerator going to 0); here, weโll instead divide everything through be the highest power in the denominator:
lim xโโโ
5 x^2 x + 3
= lim xโโโ
5 x^2 x x + 3 x
= lim xโโโ
5 x 1 + (^) x^3
Now (^3) x โ 0 as x โ โโ, so the denominator is going to 1 (which is positive). The numerator is going to โโ since we have a positive constant times x, so the entire function is going to be negative: (+)((+)โ )= (โ). Thus,
lim xโโโ
5 x^2 x + 3
Evaluate lim xโโ
e^3 โx 2
To show this one formally, we first note that as x โ โ, then
x^2 โ โ
as well, so โx^2 โ โโ
and 3 โ x^2 โ โโ
also. So, we can replace the โ3 โ x^2 โ in the exponent with another variable (say, t) that goes to โโ without changing the limit, i.e.,
lim xโโ
e^3 โx 2 = lim tโโโ
et^ (= 0 by properties mentioned in class).
Evaluate
lim xโโโ
ex
In this example, we first rewrite the limit as
lim xโโโ eโx,
which is +โ from properties mentioned in class.