Question Video: Finding the Limit of a Function Mathematics

Determine limit_(π‘₯ ⟢ 2) ((1/π‘₯⁡) βˆ’ (1/32))/(π‘₯Β² βˆ’ 4).

03:07

Video Transcript

Determine the limit as π‘₯ approaches two of one divided by π‘₯ to the fifth power minus one divided by 32 all divided by π‘₯ squared minus four.

In this question, we’re asked to evaluate a limit. And we can see the function we’re asked to evaluate the limit of is constructed of rational functions. This means we can attempt to evaluate this limit by direct substitution. To do this, we substitute π‘₯ is equal to two into our function. We get one divided by two to the fifth power minus one over 32 all divided by two squared minus four. However, if we evaluate this expression, we end up with zero divided by zero. This is an indeterminate form, which means we can’t evaluate this limit by using direct substitution alone. Instead, we’re going to need to use some other method.

To evaluate this limit, we need to notice that we can rewrite our limit. First, by using the laws of exponents, we know we can rewrite one over π‘₯ to the fifth power as π‘₯ to the power of negative five. In fact, we can also write the second term in our numerator as a power of negative five. One over 32 is two to the power of negative five. This then gives us the limit as π‘₯ approaches two of π‘₯ to the power of negative five minus two to the power of negative five all divided by π‘₯ squared minus four.

And now, this is almost exactly in the form of the limit of a difference of two powers. We just need to rewrite our denominator. We can see that four is equal to two squared. The base of this expression is the value we’re taking the limit to, and the exponent of this expression is the exponent of π‘₯ in our denominator. This gives us the limit as π‘₯ approaches two of π‘₯ to the power of negative five minus two to the power of negative five all divided by π‘₯ squared minus two squared.

And now, we can evaluate this limit by recalling a result involving the limit of a difference of powers. And we recall this tells us for any real constants π‘Ž, 𝑛, and π‘š, the limit as π‘₯ approaches π‘Ž of π‘₯ to the 𝑛th power minus π‘Ž to the 𝑛th power all divided by π‘₯ to the power of π‘š minus π‘Ž to the power of π‘š is equal to 𝑛 divided by π‘š multiplied by π‘Ž to the power of 𝑛 minus π‘š. And this is true provided the value of π‘š is not equal to zero and π‘Ž to the 𝑛th power, π‘Ž to the π‘šth power, and π‘Ž to the power of 𝑛 minus π‘š all exist.

We can see from our expression the exponent in the numerator 𝑛 is negative five, the exponent in the denominator π‘š is two, and our value of π‘Ž is also equal to two. Therefore, we can evaluate this limit by substituting these values into our formula. We get negative five divided by two multiplied by two to the power of negative five minus two. And now, we can simplify this expression negative five minus two is negative seven. And using the laws of exponents, two to the power of negative seven is one divided by two to the seventh power, which gives us negative five over two multiplied by one divided by two to the seventh power which, if we evaluate, is equal to negative five divided by 256.

Therefore, we were able to show the limit as π‘₯ approaches two of one over π‘₯ to the fifth power minus one over 32 all divided by π‘₯ squared minus four is equal to negative five divided by 256.

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