Function Limit Exercises, Limits on Intervals and Limits at a Specific Point

Let's consider an arbitrary . We must prove that there exists a that satisfies the definition of limit.

 

We wish that when
<



(which we don't know yet), it holds that:

 

 

If we simplify a bit, we have:

 

 

Note that we are trying to isolate


on one side of the inequality so we can obtain an expression for



If we multiply both sides of the inequality by 2, we will get:

 

<

 

Therefore, if we take


the inequality is satisfied. With this we finish the proof.

 

What does this mean?

 

 

It means that for any that we are given, we can always find a that satisfies the definition because we already have the formula for it— in this case.

 

For example, if we had
,

then

.

Therefore, if
  <

is equivalent to

 

< < .

 

That is,
.

Therefore, if we take any in the interval it will hold that

 

  <

 

For example, let's take
,

then
.

This way:

 

 

Similarly, if we take
,

Then
. 

Therefore:

 

1 The first limit we are asked for is

 

 

Notice that when , that is, when decreases "infinitely," the function also decreases infinitely. Therefore,

 

 

2 Now we are asked to determine

 

 

In this case let's observe that when approaches very closely from the left, the function decreases infinitely. On the other hand, when approaches very closely from the right, the function increases infinitely. Therefore, we can conclude that the limit does not exist.

 

3 Now we are asked to calculate

 

 

which refers to the left-hand limit. Let's observe that when approaches from the left, the function decreases infinitely. Therefore, the limit is

 

 

 

4 The fourth limit is

 

 

which is now the right-hand limit. Therefore,

 

 

5 Finally we are asked for the limit

 

 

which is very similar to the first limit. We can observe that as grows indefinitely, also grows indefinitely. Therefore,

 

 

We want to find the limit

 

 

Notice, first, that if we "evaluate at infinity," we obtain an indeterminate form:

 

 

 

Since the value of is not determined, we need to perform an algebraic manipulation of our function.

 

 

Before doing the algebraic manipulation, let's transform the limit using the property:

 

 

 

With which the limit becomes:

 

 

 

Now we need to algebraically manipulate the limits in order to eliminate the subtraction of infinities. This is achieved by "rationalizing" (that is, multiplying and dividing by the conjugate):

 

 

 

Notice that if we evaluate at infinity, we again have a new indetermination. In this case it's an indetermination . To get rid of this indetermination we must perform another algebraic manipulation. In this case it's multiplying and dividing by :

 

 

 

Which is the result we were looking for.

 

Now we have the limit:

 

 

Notice that if we evaluate at infinity directly, we obtain:

 

 

 

To get rid of this indetermination we can add the fractions (using the common denominator):

 

 

 

Since we now have an indetermination of the form , we will look to multiply and divide by an appropriate term to avoid the indetermination. The appropriate term is the monomial of highest degree in the numerator or denominator, that is, . Therefore, we have:

 

 

So the limit is .

 

If we evaluate at infinity, we obtain the following:

 

 

This is an indetermination of . Therefore, we must multiply and divide by the monomial of highest degree (either in numerator or denominator). In this case it's —since the denominator involves a cube root, then the "monomial" is considered as a monomial of degree 1—. Thus:

 

 

 

When we "evaluate at infinity" we have:

 

 

If we evaluate the function at infinity, we obtain:

 

 

Therefore, being an indeterminate form, we must multiply and divide by the monomial of highest degree (; remember that we divide the powers of the numerator by 2 since they are inside a square root):

 

 

 

Then, evaluating at infinity, we have:

 

 

 

Therefore, the limit is 2.

If we evaluate the function at infinity, we can immediately see that:

 

 

Therefore, we must perform an algebraic manipulation to get rid of the indetermination.

This limit can be solved in two ways.

1 By comparison of infinities:

 

First we expand the binomial squared, thus we obtain

 

 

Notice that in the numerator we obtain while in the denominator we have as the highest degree terms. Consequently, the degree of the numerator is greater and the limit will be . Remember that when the degree of the numerator is greater, then the limit is if the highest degree terms have the same sign.

 

 

2 Dividing numerator and denominator by the highest degree term:

 

 

Notice that, in general, is an indeterminate form. However, we are calculating a limit when , furthermore, both the numerator and denominator are positive as grows. Therefore, we can conclude that the limit will be .

 

If we evaluate at infinity, we have:

 

 

However, the infinity in the denominator has a higher order. Therefore, we can conclude that .

 

Nevertheless, proving that the limit is 0 without using L'Hôpital or the "order" criterion is complicated. To do it, let's denote:

 

 

To find the limit, we must look for two functions and such that and

 

 

If we find these functions, then we can conclude that .

 

First, let's observe that when , then (and this holds when is large). Similarly, we have that for . Therefore, we have that:

 

 

when "is sufficiently large." Thus, we have that where it's clear that .

 

 

To find the second function, first notice that is an increasing function, therefore, since , then . Thus, we have that

 

 

 

now, if we take , then we can write

 

 

 

A very important property about the exponential function is

 

 

 

for any and . If we take , then we have

 

 

 

From this it follows that

 

 

 

Therefore, we have that

 

 

 

Thus,

where
.

Furthermore,
.

Therefore,
.

 

If we evaluate at , then we obtain

 

 

Here we can also observe the "order" of the denominator and numerator to determine the limit. In this case the numerator has a limit of higher order, therefore,

 

 

 

To prove it, let's denote

 

and we must find a function such that

and
.

First, let's observe that when
,

then



(that is, the radicand is positive for a sufficiently large ).

Furthermore,



when

,

therefore

 

Now notice that if , then . From this it follows that

 

 

 

Therefore, . Similarly, since we know that

 

 

 

when x > 0, then we have that

 

 

 

Thus, we have that

 

 

 

where . Therefore, the limit is

 

 

When evaluating at infinity, we have that

 

 

Therefore, the limit must be infinity, since both the numerator and denominator are positive for a sufficiently large . To see it more clearly, notice that

 

 

 

Thus,

 

If we evaluate at 0, we have

 

 

where we have an indeterminate form of the type . Notice that in this case we cannot say that since we are calculating a limit when and is positive or negative near 0 (when calculating limits this is usually not a problem).

 

 

Therefore, we must calculate the one-sided limits:

 

 

1 First we calculate the left-hand limit

 

 

2 Now we calculate the right-hand limit

 

 

Here we can observe that the two one-sided limits are different, therefore, the limit when does not exist.

If we "evaluate at infinity," we obtain

 

 

Therefore, we must perform an algebraic manipulation. First let's write the limit as a single fraction:

 

 

 

Now let's divide the numerator and denominator by the highest degree coefficient (that is since it's inside a square root):

 

 

 

Now we can evaluate at infinity, so we obtain:

 

 

 

Therefore, the limit is .

 

If we evaluate at 0, we can observe that

 

 

so we have an indetermination and need to perform an algebraic manipulation. For this, we expand the binomial squared:

 

 

 

If we evaluate now at 0, we obtain:

 

 

 

Therefore, the limit is 2.

If we evaluate at 3, we will obtain

 

 

So we can observe that we have, again, an indeterminate form. To avoid the indetermination we will factor both the numerator and denominator (observe that the numerator is a difference of squares; while the denominator can be factored by finding a pair of numbers that when multiplied give 6 and when added give -5):

 

 

 

In this way, we can cancel to have

 

 

 

Thus, the limit is 6.

If we evaluate the function at 3, we observe that

 

 

So we have an indetermination and must perform an algebraic manipulation. In this case, since we have an expression with radicals, we must multiply and divide the fraction by the conjugate of the numerator (since it's the one that has the radical):

 

 

 

which is obtained because we have "conjugate binomials" in the numerator. Simplifying:

 

 

 

Now we can evaluate at 3, from which we obtain:

 

 

 

Therefore, the limit is .

 

If we evaluate at , we will obtain:

 

 

This is an indeterminate form. Therefore, we must perform algebraic manipulations to obtain some expression similar to the definition of . To do this, we first add and subtract 2 in the exponent (we will denote the limit as ):

 

 

 

Then we use properties of exponents:

 

 

Notice that the first power coincides with the definition of with . Therefore, the limit is:

 

When evaluating at infinity, we have:

 

 

Therefore, we must perform an algebraic manipulation to obtain an expression similar to the definition of . To do this, we first modify the expression we have inside the parentheses (and denote the limit as ):

 

 

 

Therefore, in the exponent we must have somehow. Notice that this is achieved by doing:

 

 

 

Then we use properties of exponents:

 

 

 

Now we can calculate the limit (noting that what's inside the square brackets is the definition of with ):

 

 

If we evaluate at , we immediately see that we have:

 

 

To solve it, we will use comparison of infinities in the two fractions. First notice that

 

 

 

since the numerator is a polynomial of higher degree (and the leading coefficients have the same sign). Similarly,

 

 

 

for the same reason. Therefore, the limit is:

 

 

 

since the last expression is not an indeterminate form.

 

Again, if we evaluate at infinity we will obtain:

 

 

However, if we solve the limits of the fractions (comparing the orders of the infinities), we can observe that:

 

 

 

since the numerator is a polynomial of higher degree and the leading coefficients have the same sign. On the other hand,

 

 

 

because the numerator is a polynomial of higher degree, but the leading coefficients have different signs. In this way, we have that the limit is:

 

 

Again, if we evaluate at infinity we have that

 

 

Therefore, we evaluate the limits of the fractions separately. In the base of the power we have that the numerator is a polynomial of higher degree, in addition to the leading coefficients having the same sign, therefore

 

 

 

On the other hand, in the exponent both the numerator and denominator are polynomials with the same order. Consequently, the limit is the quotient of the leading coefficients:

 

 

 

Therefore, the limit is

 

 

 

since if .

If we evaluate at infinity we have that

 

 

So we have to evaluate the limits of the fractions separately to determine the general limit. We have that

 

 

 

since the denominator is a polynomial of higher degree. Similarly,

 

 

 

because the numerator is a polynomial of higher degree and the leading coefficients have the same sign. Therefore, the limit is

 

 

 

because (that is, it's not an indeterminate form).

 

This limit is practically the same as the previous one, with only one sign difference.

 

 

So we have to evaluate the limits of the fractions separately to determine the general limit. Just like in the previous case, we have that

 

 

 

since the denominator is a polynomial of higher degree. Similarly, we have that

 

 

 

because the numerator is a polynomial of higher degree and the leading coefficients have different signs. Therefore, the limit is

 

 

 

notice that this is an indeterminate form. Therefore, we must use the property:

 

 

 

so

 

 

In this case,

 

 

So

 

 

Note: it's important to write it this way. The reason is that is an indeterminate form (that is, we can construct different functions where that indeterminate form "converges" to a different value).

 

Just like in the previous cases, if we evaluate at infinity, we obtain

 

 

And just like in the previous cases, we calculate the limits of each fraction separately:

 

 

 

since the numerator and denominator are polynomials of the same degree (so the limit is the quotient of the leading coefficients). Similarly, we have that

 

 

 

so the limit is

 

As in the previous cases, when evaluating at infinity we have

 

 

We calculate the limits of each fraction separately:

 

 

 

since the numerator and denominator are polynomials of the same degree (so the limit is the quotient of the leading coefficients). Furthermore, we have that

 

 

 

since the numerator is a polynomial of higher degree and the leading coefficients have different signs. Thus, the limit is

 

 

 

since we have an expression of the form where r > 1.

 

As in the previous cases, when evaluating at infinity we have

 

 

note that the function is practically the same as the previous one, but with only one sign changed. We calculate the limits of each fraction separately:

 

 

 

since the numerator and denominator are polynomials of the same degree (so the limit is the quotient of the leading coefficients). Furthermore, we have that

 

 

 

because the numerator is a polynomial of higher degree and the leading coefficients have the same sign. In this way, the limit is

 

 

 

since we have an expression of the form where 0 < r < 1.