Video Transcript
Find the limit as 𝑥 tends to zero of cot of 𝜋 by four plus 𝑥 minus one all over 𝑥.
Let’s try to evaluate this limit using direct substitution. Substituting 𝑥 equals zero, we obtain cot of 𝜋 by four minus one all over zero. Now, something divided by zero is undefined. And so, the method of direct substitution will not work as a way of evaluating the limit. However, if we find that the numerator also equals zero, then we would have obtained an indeterminate form. This would indicate that we can use L’Hopital’s rule in an attempt to evaluate the limit.
L’Hopital’s rule says that if the limit as 𝑥 tends to 𝑎 of the quotient 𝑓 of 𝑥 over 𝑔 of 𝑥 is of the indeterminate form zero over zero, ∞ over ∞, or negative ∞ over ∞, then the limit as 𝑥 tends to 𝑎 of the quotient 𝑓 of 𝑥 over 𝑔 of 𝑥 is equal to the limit as 𝑥 tends to 𝑎 of the quotient followed by the first derivative of 𝑓 over the first derivative of 𝑔. So let’s see if we can obtain zero in the numerator of our expression.
Since cot is just a shorthand notation for one over tan. We obtain one over tan of 𝜋 by four minus one all over zero. In order to work out tan of 𝜋 by four, recall the special right-angled triangle with hypotenuse square root of two and both other sides of length one. In this triangle, the two angles other than the right angle are 𝜋 over four radians. Using the ratio tan of 𝜃 equals opposite over adjacent, we can deduce from the special triangle that tan of 𝜋 by four equals one over one, which is just one. Substituting tan of 𝜋 by four equals one into the numerator of our expression, we obtain one over one minus one all over zero, which equals zero over zero. So we do, in fact, obtain the indeterminate form zero over zero when we try to evaluate the limit using direct substitution.
Now, let’s try to use L’Hopital’s rule in an attempt to evaluate the limit. Let 𝑓 of 𝑥 equal cot of 𝜋 by four plus 𝑥 minus one and 𝑔 of 𝑥 equal 𝑥. Then, the limit we require is the limit as 𝑥 tends to zero of 𝑓 of 𝑥 over 𝑔 of 𝑥, which is equal to the limit as 𝑥 tends to zero of the first derivative of 𝑓 over the first derivative of 𝑔 by L’Hopital’s rule. So we need the first derivatives of each of the functions 𝑓 and 𝑔 in order to proceed further.
For our constant 𝑐, the derivative of cot of 𝑐 plus 𝑥 with respect to 𝑥 is negative cosec squared of 𝑐 plus 𝑥. We can either memorize this result or use the quotient rule to deduce it. Using this result, the derivative of 𝑓 with respect to 𝑥 is negative cosec squared of 𝜋 by four plus 𝑥. Here, we have also used the result that the derivative of the constant negative one is zero. The derivative of a constant multiplied by 𝑥 with respect to 𝑥 is just that constant. Using this result, the derivative of 𝑔 with respect to 𝑥 is one. So using L’Hopital’s rule, we obtain that the limit in question is equal to the limit as 𝑥 tends to zero of negative cosec squared of 𝜋 by four plus 𝑥 over one. Since cosec is just shorthand notation for one over sin, we can rewrite the limit as the limit as 𝑥 tends to zero of negative one over sin squared of 𝜋 by four plus 𝑥.
Now that we have written the limit in a different form, let’s try to evaluate it using direct substitution. Substituting 𝑥 equals zero, we obtain negative one over sin squared of 𝜋 by four. Using the ratio sin of 𝜃 equals opposite over hypotenuse and the special triangle mentioned earlier, we deduce that sin of 𝜋 by four is equal to one over square root of two. Substituting this into the expression of the limit, we obtain that the limit is equal to negative one over the square of one over square root of two. This simplifies to negative one over a half which is equal to negative two.
So we obtain that the limit in question is equal to negative two.
