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.
Consider the following limit.
Look at the highest exponent of x in the above rational function. The highest exponent of x is 3.
Factor x^{3} out in both numerator and denominator. Then, simplify and evaluate the limit.
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, x^{4} 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 x^{3}, if x takes a negative value, the value of x^{3} also will be a negative value. Because, the exponent in x^{3} is 3, which is an odd number.
For example,
(-2)^{3} = -2 x -2 x -2
(-2)^{3 }= -8
When x approaches a very large negative value, x^{3} also will approach a very large negative value.
Therefore,
Problem 3 :
Solution :
Since the exponent in x^{5 }is 5, which is an odd number, when x approaches a very large negative value, x^{5} 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 e^{x}, since e > 1, when x approaches a very large positive value, e^{x }also will approach a very large positive value.
Problem 5 :
Solution :
Problem 6 :
Solution :
Problem 7 :
Solution :
Problem 8 :
Solution :
Problem 9 :
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Problem 10 :
Solution :
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