Question Video: Finding the Limit of a Difference of Powers | Nagwa Question Video: Finding the Limit of a Difference of Powers | Nagwa

Question Video: Finding the Limit of a Difference of Powers Mathematics • Second Year of Secondary School

Find lim_(𝑥 → 0) ((𝑥 − 4)² − 16)/𝑥.

02:36

Video Transcript

Find the limit as 𝑥 tends to zero of 𝑥 minus four squared minus 16 over 𝑥.

The first thing to try here is direct substitution: can we just plug zero in to the expression? We replace 𝑥 by zero and get zero minus four squared minus sixteen all over zero. Zero minus four is negative four and negative four squared is 16. So simplifying further, we find that it’s just directly substituting zero into the expression, we get the indeterminate form zero over zero.

We’re going to have to be more clever here. We need to simplify this expression first before substituting in. The first thing we do is to distribute. 𝑥 minus four squared becomes 𝑥 squared minus eight 𝑥 plus 16. We can then cancel the plus 16 and the minus 16. So we’re left with just 𝑥 squared minus eight 𝑥 in the numerator.

Both terms in the numerator have a factor of 𝑥. And so we can factor the numerator into 𝑥 times 𝑥 minus eight. And the factor of 𝑥 in the numerator cancels with the 𝑥 in the denominator, leaving us with just 𝑥 minus eight. As these two expressions are equal, their limits must be equal. And while we saw that’s direct substitution on the left-hand side gave the indeterminate form zero over zero, direct substitution on the right-hand side — plugging in 𝑥 equals zero — gives negative eight. And hence, the value of the limit that we’re looking for is negative eight.

A reasonable question to ask now is if these two expressions are supposed to be equal, then why did plugging in the zero to the left-hand side give a different answer to plugging in zero to the right-hand side? On the left-hand side, we have the indeterminate form zero over zero and on the right-hand side, we had negative eight. The answer is that in our last step of algebra, where we cancel the factor of 𝑥 in the numerator with the 𝑥 in the denominator, we turn something which is undefined when 𝑥 is zero into something which is defined.

The two expressions are equal for all nonzero values of 𝑥. But for 𝑥 is zero, the left-hand side expression is undefined. As we’re taking the limit as 𝑥 tends to zero, we don’t care about what the value of the expression is when 𝑥 is zero. We only care about values of 𝑥 nearby. The limits then are in fact really equal. And as we’ve seen are in fact equal to negative eight.

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