Indeterminate Form

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

To solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

For polynomials,
the fastest increasing terms
are the highest order terms.

The highest order term in the numerator is
3n².

The highest order term in the denominator is
n².

The orders of 3n² and n² are the same: 2.

Then write the coefficients of the terms:
3/1.
This is the limit value.

3/1 = 3

So 3 is the answer.

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

3^{n + 4}
= 3ⁿ⋅3⁴
= 81⋅3ⁿ

+2^{2n + 1}
= +2⋅2²ⁿ
= +2⋅4ⁿ

Product of Powers

Power of a Power

To solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

An exponential term whose base is greater than 1
is faster than any polynomial term.
And the exponential term that has the bigger base
increases faster.

So the fastest term in the numerator is
+2⋅4ⁿ.

And the fastest term in the denominator is
+4ⁿ.

The bases of +2⋅4ⁿ and +4ⁿ
are the same: 4.

Then write the coefficients of the terms:
2/1.
This is the limit value.

2/1 = 2

So 2 is the answer.

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

To solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

The highest order term in the numerator is
3n.

The highest order terms in the denominator are
√4n² ... (= 4n^2/2) and +√n² ... (= +n^2/2).

Rational Exponent

The orders of the terms are the same: 1.

Then write the coefficients of the terms:
3/(√4 + √1).
This is the limit value.

√4 = √2² = 2
√1 = 1

Square Root

2 + 1 = 3

3/3 = 1

So 1 is the answer.

As n → ∞,
√n² + 6n - 12 goes to ∞
and n goes to ∞.

So this is an indeterminate form ∞ - ∞.

To solve ∞ - ∞ form,
multiply the conjugate of (√n² + 6n - 12 - n),
(√n² + 6n - 12 + n),
to both of the numerator and the denominator.

(√n² + 6n - 12 - n)(√n² + 6n - 12 + n)
= (n² + 6n - 12 - n²)

Product of a Sum and a Difference: (a + b)(a - b)

Cancel n² and -n².

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

So, to solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

The highest order term in the numerator is
6n.

The highest order terms in the denominator are
√n² ... (= n^2/2) and +n.

The orders of the terms are the same: 1.

Then write the coefficients of the terms:
6/(√1 + 1).
This is the limit value.

√1 + 1 = 1 + 1 = 2

6/2 = 3

So 3 is the answer.

As n → ∞,
√n² + n goes to ∞
and √n² - n goes to ∞.

So the denominator is an indeterminate form ∞ - ∞.

To solve ∞ - ∞ form,
multiply the conjugate of (√n² + n - √n² - n),
(√n² + n + √n² - n),
to both of the numerator and the denominator.

(√n² + n - √n² - n)(√n² + n + √n² - n)
= n² + n - (n² - n)

Product of a Sum and a Difference: (a + b)(a - b)

-(n² - n) = -n² + n

Cancel n² and -n².
+n + n = 2n

As n → ∞,
the numerator goes to ∞
and the denominator goes to ∞.

So this is an indeterminate form ∞/∞.

So, to solve ∞/∞ form,
find the fastest increasing terms
from the numerator and the denominator.

The highest order terms in the numerator are
√n² ... (= n^2/2) and +√n² ... (= +n^2/2).

The highest order term in the denominator is
2n.

The orders of the terms are the same: 1.

Then write the coefficients of the terms:
7(√1 + √1)/2.
This is the limit value.

√1 = 1

(1 + 1) = 2

7⋅2/2 = 7

So 7 is the answer.

As x → 1,
the numerator goes to 0
and the denominator goes to 0.

So this is an indeterminate form 0/0.

To solve 0/0 form,
make a factor that makes the denominator 0, (x - 1),
and cancel the factors.

To make (x - 1),
factor x² + x - 2.

2⋅(-1) = -2
2 - 1 = +1

So x² + x - 2 = (x + 2)(x - 1).

Factor a Quadratic Trinomial

Cancel (x - 1) factors.

Find the limit value.

Put 1 into (x + 2).

Limit of a Function

1 + 2 = 3

So 3 is the answer.

As x → 2,
the numerator goes to 0
and the denominator goes to 0.

So this is an indeterminate form 0/0.

To solve 0/0 form,
make a factor that makes the denominator 0, (x - 2),
and cancel the factors.

To make (x - 2),
multiply the conjugate of (√x + 7 - 3),
(√x + 7 + 3),
to both of the numerator and the denominator.

(√x + 7 - 3)(√x + 7 + 3)
= x + 7 - 9

Product of a Sum and a Difference: (a + b)(a - b)

+7 - 9 = -2

Cancel (x - 2) factors.

Find the limit value.

Put 2
into 1/(√x + 7 + 3).

2 + 7 = 9

√9 = √3² = 3

3 + 3 = 6

So 1/6 is the answer.

As x → 0,
[1/x] goes to ∞
and [6/(x + 3) - 2] goes to 0.

So this is an indeterminate form 0⋅∞.

To solve 0⋅∞ form,
make a factor that makes the denominator 0, x,
and cancel the factors.

To make x,
simplify [6/(x + 3) - 2].

-2 = -2⋅(x + 3)/(x + 3).

Add and Subtract Rational Expresssions

6 - 2(x + 3) = 6 - 2x - 6

Cancel 6 and -6.

Cancel x factors.

Find the limit value.

Put 0
into -2/(x + 3).

0 + 3 = 3

So -2/3 is the answer.