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Limit of Rational Function...1?

Find the limit of (2x + 1)/(x + 4) as x tends to - 4 from the right side.

I know there's a vertical asymptote at x = -4. I think the best way to solve this problem is by graphing the function. I am not too sure about how to solve algebraically.

I am thinking about the number line.

<---------------(-4)----------------->

What if I select values to the left and right of -4 but not including -4? By doing this, I will then know if the interval (-00, -4) is positive or negative and if the interval (-4, 00) is positive or negative.

Is this correct so far? What's the limit? 

2 Answers

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  • 3 days ago
    Favorite Answer

    f(x) = (2x + 1)/(x + 4)

    We can re-write this as 2(x + 4)/(x + 4) - 7/(x + 4)

    i.e. 2 - [7/(x + 4)]

    As x --> ±∞, 7/(x + 4) --> 0

    Hence, the limit tends to 2

    As you say, x = -4 is a vertical asymptote, so we need to establish the behaviour of f(x) for values less than -4 and values more than -4

    If we choose x < -4, i.e. negative values, we have:

    f(-ve) = 2 - [7/(-ve)] => 2 + (+ve)

    This means the values of f(x) are greater than 2

    Hence, for values less than x = -4, f(x) approaches a limit of 2 from above

    Choosing x > -4, we have:

    2 - [7/(+ve)] => 2 - (+ve)

    This means the values of f(x) are less than 2

    Hence, for values greater than x = -4, f(x) approaches a limit of 2 from below.

    No need to sketch the graph, but it does help us to visualise the above.

    :)>

     

    Attachment image
  • 4 days ago

    The numerator approaches 2(-4) + 1 = -8 + 1 = -7

    As x approaches -4 from above ("the right side"), then x + 4 approaches 0 from above, meaning the denominator is positive.

    With the numerator negative and the denominator positive, the overall result is negative.

    With the denominator approaching zero, and the numerator NOT approaching zero, that means the overall value approaches positive or negative infinity.

    Put those together and the answer is negative infinity.

    No graphing is necessary.

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