Video: Finding the Value of a Limit Involving Exponential and Trigonometric Functions Using L’HΓ΄pital’s rule

Determine lim_(π‘₯ β†’ 0) βˆ’(6 sin 5π‘₯)/(2^(π‘₯) βˆ’ 1).

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Video Transcript

Determine the limit as π‘₯ approaches zero of negative six times the sin of five π‘₯ all divided by two to the power of π‘₯ minus one.

The question is asking us to determine the limit of the quotient of two functions. We can see the function in our numerator is made up of polynomials, compositions, and standard trigonometric functions. Similarly, the function in the denominator is an exponential function, and then we just subtract the constant one. So we could attempt to evaluate the limit of our numerator and the limit of our denominator by direct substitution.

So by substituting π‘₯ is equal to zero, we get negative six times the sin of five times zero all divided by two to the zeroth power minus one. We know the sin of zero is equal to zero, so our numerator is equal to zero. And two to the zeroth power is just equal to one, so our denominator is also equal to zero. So using direct substitution gave us an indeterminate form. This means we’re going to need to evaluate our limit in a different method.

We recall if the limit of the quotient of two functions gives us an indeterminate form, then we can try using L’HΓ΄pital’s rule. One version of L’HΓ΄pital’s rule tells us that if we have two differentiable functions 𝑓 and 𝑔, where the derivative of 𝑔 is not equal to zero for values of π‘₯ around π‘Ž, although 𝑔 prime of π‘Ž is allowed to be equal to zero. And we also have the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ are both equal to zero. Then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯.

In other words, if the limit as π‘₯ approaches π‘Ž of the quotient of two functions gives us an indeterminate form, then, under certain circumstances, we can evaluate this limit as the limit as π‘₯ approaches π‘Ž of the quotient of the derivatives of these two functions. So let’s see if we can apply this version of L’HΓ΄pital’s rule to the limit given to us in the question.

We’ll set our function 𝑓 of π‘₯ to be the function in the numerator of our limit. That’s negative six times the sin of five π‘₯. And we’ll set our function 𝑔 of π‘₯ to be the function in the denominator of our limit. That’s two to the power of π‘₯ minus one. Finally, since we want to find the limit as π‘₯ approaches zero, we’ll set our value of π‘Ž equal to zero.

We now need to check that all of the prerequisites for L’HΓ΄pital’s rule are true. First, let’s check that our functions 𝑓 and 𝑔 are differentiable. We already said that our numerator is the composition of polynomials and standard trigonometric functions. Since all of these are differentiable, their compositions will also be differentiable. So our numerator being the composition of differentiable functions must be differentiable. It’s a similar story in our denominator. We know the exponential function two to the power of π‘₯ is differentiable. We then subtract one from this and it’s still differentiable.

Next, we need to show that the derivative of 𝑔 of π‘₯ is not equal to zero around π‘₯ is equal to zero. To check that this is true, let’s find an expression for 𝑔 prime of π‘₯. We recall that 𝑔 of π‘₯ is equal to two to the power of π‘₯ minus one. There’s a few different ways of differentiating this. We could recall a standard result to differentiate exponential functions. However, we can do this directly.

We recall by using our laws of exponents and our laws of logarithms π‘Ž to the power of π‘₯ is equal to 𝑒 to the power of the natural logarithm of π‘Ž all raised to the power of π‘₯. This is because 𝑒 to the power of the natural logarithm of π‘Ž is just equal to π‘Ž. Then, by using our laws of exponents, we can rewrite this as 𝑒 to the power of π‘₯ times the natural logarithm of π‘Ž. Using this, we can rewrite our function 𝑔 of π‘₯ as 𝑒 to the power of π‘₯ times the natural logarithm of two minus one.

We now see this is in a form which we can differentiate. For any constant 𝑛, we know the derivative of 𝑒 to the power of 𝑛π‘₯ with respect to π‘₯ is equal to 𝑛 times 𝑒 to the power of 𝑛π‘₯. So we can differentiate each term in our function for 𝑔 of π‘₯ to get 𝑔 prime of π‘₯ is equal to the natural logarithm of two times 𝑒 to the power of π‘₯ times the natural logarithm of two minus zero. And of course, we can just remove the minus zero.

Remember, we’re checking that 𝑔 prime of π‘₯ is not equal to zero around π‘₯ is equal to zero. We can see that the natural logarithm of two is a positive constant. And we have 𝑒 raised to the power of any number is also a positive constant. So we have that 𝑔 prime of π‘₯ is positive. So we’ve shown that the derivative function is positive. In particular, this means that it’s not equal to zero for any value of π‘₯.

The last prerequisite we need to show is true is that the limit as π‘₯ approaches zero of 𝑓 of π‘₯ and the limit as π‘₯ approaches zero of 𝑔 of π‘₯ need to both be equal to zero. However, we’ve already shown this to be true. When we initially try to evaluate our limit by using direct substitution, the limit we evaluated in the numerator is equal to the limit as π‘₯ approaches zero of 𝑓 of π‘₯. We evaluated both of these limits by using direct substitution. And we got zero divided by zero. So in fact, we’ve already shown that both of these limits are already equal to zero. So now, we’ve shown that all of our prerequisites are true.

This means instead of evaluating the limit given to us in the question directly, we can instead try to evaluate the limit as π‘₯ approaches zero of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯. Before we do this, let’s first find an expression for 𝑓 prime of π‘₯. We recall 𝑓 of π‘₯ is equal to negative six times the sin of five π‘₯. And we can differentiate this by using one of our derivative rules for trigonometric functions. For constants π‘Ž and 𝑛, the derivative of π‘Ž times the sin of 𝑛π‘₯ with respect to π‘₯ is equal to 𝑛 times π‘Ž times the cos of 𝑛π‘₯.

Applying this, we get that 𝑓 prime of π‘₯ is equal to five times negative six times the cos of five π‘₯, which we can simplify to give us negative 30 times the cos of five π‘₯. So we’re now ready to attempt to evaluate this limit by using L’HΓ΄pital’s rule. First, this version of L’HΓ΄pital’s rule told us the limit as π‘₯ approaches zero of 𝑓 of π‘₯ divided by 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches zero of 𝑓 prime of π‘₯ divided by 𝑔 prime of π‘₯.

Next, we showed that 𝑓 prime of π‘₯ is equal to negative 30 times the cos of five π‘₯ and 𝑔 prime of π‘₯ is equal to the natural logarithm of two times 𝑒 to the power of π‘₯ times the natural logarithm of two. We can again see that both of these functions are made up of standard functions. So we can attempt to evaluate this by using direct substitution. Substituting π‘₯ is equal to zero, we get negative 30 times the cos of five times zero all divided by the natural logarithm of two times 𝑒 to the power of zero times the natural logarithm of two.

We know the cos of zero is just equal to one and zero times the natural logarithm of two is equal to zero and 𝑒 raised to the zeroth power is also equal to one. So this simplifies to give us negative 30 divided by the natural logarithm of two. Therefore, by using L’HΓ΄pital’s rule, we’ve shown that the limit as π‘₯ approaches zero of negative six times the sin of five π‘₯ all divided by two to the power of π‘₯ minus one is equal to negative 30 divided by the natural logarithm of two.

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