Video Transcript
Find the limit as 𝑥 tends to zero of 𝑥 cubed over sin cubed of two 𝑥.
If we try to use direct substitution to evaluate this limit, we obtain zero cubed, which is just zero, over sin cubed of two times zero, which is just sin cubed of zero. This equals zero over zero, using the fact that sin of zero equals zero. Now, zero over zero is undefined, which means that we cannot use direct substitution to evaluate this limit. However, zero over zero is an indeterminate form. So we can use L’Hôpital’s rule.
L’Hôpital’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 it is equal to the limit as 𝑥 tends to 𝑎 of the quotient formed by the first derivative of 𝑓 over the first derivative of 𝑔. Let’s use L’Hôpital’s rule to evaluate the limit in question.
Let 𝑓 of 𝑥 equal 𝑥 cubed and 𝑔 of 𝑥 equal sin cubed of two 𝑥. Then, we can rewrite the limit in question as the limit as 𝑥 tends to zero of 𝑓 of 𝑥 over 𝑔 of 𝑥. Using L’Hôpital’s rule, this is equal to the limit as 𝑥 tends to zero of the first derivative of 𝑓 over the first derivative of 𝑔. Let’s work out the first derivatives of the functions 𝑓 and 𝑔. The first derivative of 𝑓 is three 𝑥 squared. To obtain this, we have just used the standard formula for differentiating terms of the form 𝑎𝑥 to the power of 𝑛 with respect to 𝑥, where 𝑎 and 𝑛 are constants. The formula tells us to multiply the coefficient by the exponent and then decrease the exponent by one. Multiplying one by three, we obtain three and then decreasing three by one, we obtain two.
Since 𝑔 is composed of more than one function, we will use the chain rule to find its derivative. We can rewrite the function 𝑔 as 𝑢 cubed, where 𝑢 is the function of 𝑥 equal to sin of two 𝑥. The chain rule states that if 𝑔 is a function of 𝑢 and 𝑢 is a function of 𝑥, then their derivative of 𝑔 with respect to 𝑥 is equal to the derivative of 𝑔 with expect to 𝑢 multiplied by the derivative of 𝑢 with respect to 𝑥. In order to apply the chain rule to the function 𝑔 equals 𝑢 cubed, where 𝑢 of 𝑥 equals sin of two 𝑥, we need to find the derivative of 𝑢 cubed with respect to 𝑢 and the derivative of sin two 𝑥 with respect to 𝑥 and multiply them together.
The derivative of 𝑢 cubed with respect to 𝑢 is three 𝑢 squared, using the standard formula for differentiating terms of the form 𝑎𝑥 to the power of 𝑛. The derivative or sin two 𝑥 with respect to 𝑥 is two cos two 𝑥. We can either deduce this by separately applying the chain rule to the function sin of two 𝑥 or we can memorize the formula for the derivative of the function sin of 𝑎𝑥 with respect to 𝑥, where 𝑎 is a constant. Multiplying three 𝑢 squared by two cos two 𝑥 and then substituting sin two 𝑥 in place of 𝑢, we obtain that the derivative of 𝑔 with respect to 𝑥 is six sin squared two 𝑥 cos two 𝑥.
Therefore, using L’Hôpital’s rule, the limit in question can be rewritten as the limit as 𝑥 tends to zero of three 𝑥 squared over six sin squared two 𝑥 cos two 𝑥. We can simplify this limit by dividing the numerator and denominator by three. If we try to use direct substitution to evaluate the new rewritten limit, we again obtain the indeterminate form zero over zero. This indicates that we should use L’Hôpital’s rule for a second time.
In order to do this, let’s redefine the function 𝑓 to be 𝑥 squared and the function 𝑔 to be two sin squared two 𝑥 cos two 𝑥. In order to apply L’Hôpital’s rule, we need the derivatives of these functions. The derivative of 𝑥 squared is two 𝑥 using the standard formula for differentiating terms of the form 𝑎𝑥 to the power of 𝑛. Before we find the derivative of 𝑔, let’s rewrite 𝑔 in a slightly simpler form. Note that 𝑔 contains the term two sin two 𝑥 cos two 𝑥. Using the double angle formula sin two 𝜃 equals two sin 𝜃 cos 𝜃, we can rewrite the term two sin two 𝑥 cos two 𝑥 as sin of four 𝑥.
Having written 𝑔 after product of the terms sin four 𝑥 and sin two 𝑥, let’s use the product rule to differentiate it. The product rule says that the derivative of the product of functions 𝑢 and 𝑣 is equal to the derivative of 𝑢 multiplied by 𝑣 plus the derivative of 𝑣 multiplied by 𝑢. Letting 𝑢 equal sin four 𝑥 and 𝑣 equal sin two 𝑥 in the product rule, we obtain that the derivative of 𝑔 with respect to 𝑥 is equal to the derivative of 𝑢 with respect to 𝑥 which is four cos four 𝑥 as per the formula for the derivative of sin 𝑎𝑥 with respect to 𝑥 that we saw earlier multiplied by 𝑣, which is sin two 𝑥, plus the derivative of 𝑣 with respect to 𝑥, which is two cos two 𝑥, multiplied by 𝑢, which is sin four 𝑥.
Therefore, using L’Hôpital’s rule, we can rewrite our limit as the limit as 𝑥 tends to zero of two 𝑥 over four cos four 𝑥 sin two 𝑥 plus two cos two 𝑥 sin four 𝑥. We can simplify this limit by dividing the numerator and denominator by two. If we try to use direct substitution to evaluate this limit, once again we obtain the indeterminate form zero over zero. Once again, this indicates that we should use L’Hôpital’s rule.
Now, you might be thinking whether we are going to go on forever in this kind of loop with L’Hôpital’s rule or whether we will stop somewhere. The good news is that the third iteration of L’Hôpital’s rule is the last. This is because the derivative of the function 𝑥 in the numerator is one, using the standard formula for differentiating terms of the form 𝑎𝑥 to the power of 𝑛. Therefore, an application of L’Hôpital’s rule results in the limit in question being equal to the limit as 𝑥 tends to zero of one over the derivative of 𝑔, where 𝑔 is the function of 𝑥 equal to two cos four 𝑥 sin two 𝑥 plus cos two 𝑥 sin four 𝑥.
Since substituting 𝑥 equals zero in the constant function one does not change its value, we will not obtain any of the indeterminate forms zero over zero, ∞ over ∞, or negative ∞ over ∞, that will indicate another application of L’Hôpital’s rule after we find the derivative of 𝑔 and attempt to evaluate the limit by direct substitution. Let’s proceed further by working out the derivative of the function 𝑔 with respect to 𝑥. In order to differentiate 𝑔, we will use the product rule separately on the functions two cos four 𝑥 sin two 𝑥 and cos two 𝑥 sin four 𝑥 and add the results together.
In what follows, we need to recall that the derivative of the function cos 𝑎𝑥 with respect to 𝑥, where 𝑎 is a constant is negative 𝑎 sin of 𝑎𝑥. You can work this out using the chain rule if need be. Using the product rule, the derivative of two cos four 𝑥 sin two 𝑥 is the derivative of two cos four 𝑥, which is negative eight sin four 𝑥, multiplied by sin two 𝑥 plus the derivative of sin two 𝑥, which is two cos two 𝑥, multiplied by two cos four 𝑥. Let’s simplify this by multiplying together the two twos in the second term.
In a similar way, using the product rule, we obtain that the derivative of cos two 𝑥 sin four 𝑥 is negative two sin two 𝑥 sin four 𝑥 plus four cos four 𝑥 cos two 𝑥. Collecting like terms together, we obtain that the derivative of 𝑔 is equal to negative 10 sin two 𝑥 sin four 𝑥 plus eight cos four 𝑥 cos two 𝑥. So by L’Hôpital’s rule, the limit in question is equal to the limit as 𝑥 tends to zero of one over negative 10 sin two 𝑥 sin four 𝑥 plus eight cos four 𝑥 cos two 𝑥.
Now, let’s use direct substitution to evaluate the limit. Substituting 𝑥 equals zero into the limit and noting that sin of zero equals zero and cos of zero equals one, we obtain that the limit in question is equal to one over zero plus eight, which is just one over eight. So our final answer is one over eight.
