Question Video: Solving Limits by Transforming Them into the Natural Exponent Limit Forms Mathematics

Determine lim_(π‘₯ β†’ 0) (βˆ’4 tanΒ³ π‘₯ + 1)^(cotΒ³ π‘₯).

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Video Transcript

Determine the limit as π‘₯ approaches zero of negative four tan cubed of π‘₯ plus one all raised to the power of cot cubed of π‘₯.

In this question, we’re asked to evaluate a limit. And the first thing we should do whenever we’re asked to evaluate a limit is see if we can do this directly. We might want to try direct substitution. However, remember, trigonometric functions are only continuous on their domain. And π‘₯ is equal to zero is not in the domain of cot of π‘₯. So we can’t evaluate this by using direct substitution.

However, we can see what happens to our output as π‘₯ is getting closer to zero. Inside our parentheses, as the values of π‘₯ are getting closer to zero, the tan of π‘₯ is also getting closer to zero. So the expression inside our parentheses is getting closer to one. However, we can’t determine what happens to our exponent as π‘₯ approaches zero. As π‘₯ approaches zero from the right, cot of π‘₯ approaches ∞. However, as π‘₯ approaches zero from the left, cot of π‘₯ approaches negative ∞. And in either case, one to the power of positive ∞ and one to the power of negative ∞ are indeterminate forms anyway. So we can’t use this to evaluate our limit.

Instead, we’re going to need to try some form of manipulation. And it can be difficult to see what we need to do to help us evaluate this limit. One thing we can notice is cot cubed of π‘₯ is actually one divided by tan cubed of π‘₯. So we’ll try using this to help us evaluate our limit. We recall this tells us the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 is equal to 𝑒. And it’s worth pointing out here we can use our other limit result involving Euler’s constant 𝑒 to answer this question.

However, usually, one of the two limit results involves a lot easier working. And it can be very difficult to determine which of our two limit results we should use just by looking at the limit we’re asked to evaluate. So if we’re struggling, we should try switching limit results.

Now, to use this limit to help us evaluate this question, we’re going to need to start by rewriting our limit in this form. The first thing we’ll do is switch the order of the two terms inside our parentheses. This gives us the limit as π‘₯ approaches zero of one minus four tan cubed of π‘₯ all raised to the power of cot cubed of π‘₯. Next, inside our parentheses, we’re going to want one plus our variable. So we’re going to need to use a substitution. We’re going to substitute 𝑛 is equal to negative four times the tan cubed of π‘₯.

So to rewrite our limit by using the substitution, we need to see what happens to our values of 𝑛 as π‘₯ approaches zero. We can do this directly from our substitution. On the right-hand side of our equation, as the values of π‘₯ approach zero, the right-hand side of our equation is approaching zero, because the right-hand side of this equation is a continuous function. So we can actually just evaluate this by direct substitution. So as our values of π‘₯ are approaching zero, our values of 𝑛 are also approaching zero.

But there’s one more instance of π‘₯ which appears in our limit. We need to find an expression for cot cubed of π‘₯ in terms of 𝑛. And to do this, we’re going to need to rearrange our substitution. We start by dividing through by negative four. We get negative 𝑛 over four is equal to tan cubed of π‘₯.

And now, because we want an expression for cot cubed of π‘₯, we can actually just take the reciprocal of both sides of this equation. And it’s worth pointing out here we don’t need to worry about dividing by zero because our limits are as π‘₯ approaches zero and 𝑛 approaches zero. So our values of π‘₯ and 𝑛 are never equal to zero. And this means tan cubed of π‘₯ is not equal to zero and 𝑛 is not equal to zero. So this gives us that negative four over 𝑛 is equal to cot cubed of π‘₯.

Now we’re ready to use this substitution to rewrite our limit. We’ve rewritten our limit as the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of negative four over 𝑛. And now this is almost in exactly our form of using our limit result involving Euler’s constant 𝑒. The only difference is, in our exponent, we have a factor of negative four. So to apply our limit result, we’re going to want to take this exponent factor of negative four outside of our limit. To do this, we’re going to need to use the power rule for limits. But first, we’re going to use our laws of exponents.

We know π‘Ž to the power of 𝑏 times 𝑐 is equal to π‘Ž to the power of 𝑏 all raised to the power of 𝑐. By setting π‘Ž equal to one plus 𝑛, 𝑏 equal to one over 𝑛, and 𝑐 equal to negative four, we can rewrite our limit as the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 all raised to the power of negative four. Now, all that’s left to do is take this exponent of negative four outside of our limit. And to do this, we’re going to need to use the power rule for limits.

One version of this tells us the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 raised to the π‘˜th power is equal to the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 all raised to the π‘˜th power, provided the limit as 𝑛 approaches π‘Ž of 𝑓 of 𝑛 exists and we can raise this to the π‘˜th power. And the easiest way to show that these two prerequisites are true is to write the next step in our line of working. By using the power rule for limits, we get that our limit is equal to the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 all raised to the power of negative four.

But remember, this is provided the limit as 𝑛 approaches zero of one plus 𝑛 all raised to the power of one over 𝑛 exists, and we know this does exist. It’s our limit result involving Euler’s constant 𝑒. And of course, don’t forget we also need to check that we can raise this number to our exponent, and of course we can. We just get 𝑒 to the power of negative four. This means we were able to justify our use of the power rule for limits to get that our limit evaluates to 𝑒 to the power of negative four. However, we’ll do one last piece of manipulation to rewrite this as one over 𝑒 to the fourth power, which is our final answer.

Therefore, we were able to show the limit as π‘₯ approaches zero of negative four tan cubed of π‘₯ plus one all raised to the power of cot cubed of π‘₯ is equal to one over 𝑒 to the fourth power.

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