# EVALUATING LIMITS AT INFINITY

Consider the following limit.

When x approaches -, x < 0.

Find the value of ¹⁄ₓ, when x < 0.

x = -1 ----> ¹⁄₋₁ = -1

x = -2 ----> ¹⁄₋₂ = -0.5

x = -3 ----> ¹⁄₋₃ = -0.3

x = -4 ----> ¹⁄₋₄ = -0.25

x = -5 ----> ¹⁄₋₅ = -0.2

x = -10 ----> ¹⁄₋₁₀ = -0.1

x = -100 ----> ¹⁄₋₁₀₀ = -0.01

x = -1000 ----> ¹⁄₋₁₀₀₀ = -0.001

x = -10000 ----> ¹⁄₋₁₀₀₀₀ = -0.0001

x = -100000 ----> ¹⁄₋₁₀₀₀₀₀ = -0.00001

When x approaches -¹⁄ₓ approaches 0.

Therefore,

When x approaches +, x > 0.

When x approaches +, if the above calculation is done, ¹⁄ₓ will approach 0.

Conclusion :

When you divide any number by either negative infinity or positive infinity, the result is always zero.

## Evaluating Limits for Rational Functions at Infinity

Consider the following limit.

Look at the highest exponent of x in the above rational function. The highest exponent of x is 3.

Factor x3 out in both numerator and denominator. Then, simplify and evaluate the limit.

## Solved Problems

Problem 1 :

Solution :

x approaches positive infinity means, x becomes larger and larger. That is, x approaches a very large positive value.

When x approaches a very large positive value, x4 will approach a very large positive value. When a very large positive value multiplied by 3, again it will become a very large positive value.

Therefore,

Problem 2 :

Solution :

x approaches negative infinity means, x becomes smaller  and smaller. That is, x approaches a very large negative value.

In x3, if x takes a negative value, the value of x3 also will be a negative value. Because, the exponent in x3 is 3, which is an odd number.

For example,

(-2)3 = -2 x -2 x -2

(-2)= -8

When x approaches a very large negative value, x3 also will approach a very large negative value.

Therefore,

Problem 3 :

Solution :

Since the exponent in xis 5, which is an odd number, when x approaches a very large negative value, x5 also will approach a very large negative value.

When a very large negative value is multiplied by -3, it will become a very large positive value.

Therefore,

Problem 4 :

Solution :

e is a mathematical constant and its value is

e = 2.71828......

In ex, since e > 1, when x approaches a very large positive value, ealso will approach a very large positive value.

Problem 5 :

Solution :

Problem 6 :

Solution :

Problem 7 :

Solution :

Problem 8 :

Solution :

Problem 9 :

Solution :

Problem 10 :

Solution :

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