Video: Finding the Limit of a Quotient of a Difference of Powers at a Point

Determine lim_(π‘₯ β†’ 2) (π‘₯⁡ βˆ’ 32)/(π‘₯ βˆ’ 2).

02:53

Video Transcript

Determine the limit as π‘₯ tends to two of π‘₯ to the power of five minus 32 over π‘₯ minus two.

The first thing we can do to try to evaluate this limit is just to directly substitute the limit point two into the expression. The expression is π‘₯ to the power of five minus 32 over π‘₯ minus two. Substituting two into this, we get two to the power of five minus 32 over two minus two. Two to the power of five is 32, so we get 32 minus 32 over two minus two. And this is zero divided by zero. This is an indeterminate form, which we can’t evaluate. And so direct substitution hasn’t given us the value of this limit.

Clearly, we’re not going to have any success. Well, π‘₯ minus two is the denominator of this fraction. So we’re going to have to rewrite this into another form, where that’s not the case. If you know about the factor theorem, you’ll know that we have good reason to expect that the numerator π‘₯ to the power of five minus 32 has a factor of π‘₯ minus two. And so if we perform a polynomial long division to this rational expression, we should expect to get a polynomial.

An alternative to performing the polynomial long division is to rewrite the 32 as two to the power of five. And then we can apply the identity π‘₯ to the power of 𝑛 minus π‘Ž to the power of 𝑛 over π‘₯ minus π‘Ž equals π‘₯ to the power of 𝑛 minus one plus π‘Ž times π‘₯ to the power of 𝑛 minus two plus π‘Ž squared times π‘₯ to the power of 𝑛 minus three, and so on. So the last two terms are π‘Ž to the power of 𝑛 minus two times π‘₯ plus π‘Ž to the power of 𝑛 minus one.

So setting π‘Ž equal to two and 𝑛 equal to five, we get that this quotient is π‘₯ to the power of four plus two times π‘₯ to the power of three plus two squared times π‘₯ squared plus two to the power of three times π‘₯ plus two to the power of four. And we can simplify this by evaluating the powers of two. The left-hand side and right-hand side are equal. Apart from that, π‘₯ equals two and the left-hand side is undefined. So the limit of the left-hand side is equal to the limit of the right-hand side. And when we substitute it directly into the limits on the right-hand side, we don’t get an indeterminate form.

Replacing π‘₯ by two, we get two to the power of four plus two times two to the power of three plus four times two to the power of two plus eight times two plus 16. And evaluating this, we get our final answer 80. So to recap what happened, substituting directly into π‘₯ to the power of five minus 32 over π‘₯ minus two gave us an indeterminate form. But by applying an identity or using polynomial long division, we could rewrite this expression as a polynomial. And substituting directly into that polynomial gave us the value of the limit 80.

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