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

Determine lim_(π‘₯ β†’ 0) (π‘₯ + 1)^(11/10π‘₯).

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Video Transcript

Determine the limit as π‘₯ approaches zero of π‘₯ plus one all raised to the power of 11 over 10π‘₯.

We’re asked to evaluate a limit, and we can try and do this directly. First, we can see our values of π‘₯ are approaching zero. This means inside of our parentheses, one plus π‘₯ is approaching one plus zero, which is one. However, we then see a problem when we try to evaluate our exponent. The numerator remains constant. However, as π‘₯ is approaching zero, the denominator is approaching zero, so the size of our exponent is growing without bound. This means by trying to evaluate our limit, we get an indeterminate form. We’re going to need to try some other method of evaluating this limit.

To do this, we can take a closer look to the limit given to us in the question. It’s actually very similar to a limit which we do know how to evaluate. We can see the limit given to us in the question is very similar to a useful limit result, the limit as π‘₯ approaches zero of one plus π‘₯ all raised to the power of one over π‘₯ is equal to Euler’s constant 𝑒. We’re going to want to rewrite the limit given to us in the question in a form where we can use our limit result.

The first thing we’re going to want to do is reorder the two terms inside of our parentheses. This is just so it matches the limit result we’ve used. This gives us the limit as π‘₯ approaches zero of one plus π‘₯ all raised to the power of 11 over 10π‘₯. And now, we can see that this limit is very similar to our limit result. However, in our limit result, the exponent is one over π‘₯. However, in our limit, the exponent is 11 divided by 10π‘₯. So, we want to rewrite our exponent in terms of one over π‘₯. To do this, we’re going to start by writing 11 over 10π‘₯ as one over π‘₯ multiplied by 11 over 10. This gives us the limit as π‘₯ approaches zero of one plus π‘₯ all raised to the power of one over π‘₯ times 11 over 10.

Now, we want to write this in terms of our limit result. To do this, we need to do two things. First, we’re going to need to use our laws of exponents. First, we need to recall that π‘Ž to the power of 𝑏 times 𝑐 can be rewritten as π‘Ž to the power of 𝑏 all raised to the power of 𝑐. We can use this to rewrite our limit with π‘Ž set to be one plus π‘₯, 𝑏 set to be one over π‘₯, and 𝑐 set to be 11 over 10. Doing this, we get the limit as π‘₯ approaches zero of one plus π‘₯ all raised to the power of one over π‘₯ all raised to the power of 11 over 10. But we still can’t use our limit result because we’re raising this to the power of 11 over 10. We want to take this outside of our limit.

And in fact, we can do this in this case by using the power rule for limits. We recall this tells us for constant π‘Ž, rational number 𝑛, and function 𝑓 of π‘₯, the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ raised to the 𝑛th power is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ all raised the 𝑛th power. And this version of the power rule for limits will work, provided the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ exists and is equal to a nonnegative number.

And in fact, this is exactly the case we have. We know the limit as π‘₯ approaches zero of one plus π‘₯ all raised to the power of one over π‘₯ is equal to Euler’s constant 𝑒. So, by using the power rule for limits, we can take 11 over 10 outside of our limit. And this just means we can evaluate the limit inside of our parentheses. We know this is equal to Euler’s constant 𝑒. So, by writing this limit as 𝑒, we were able to show the limit as π‘₯ approaches zero of π‘₯ plus one all raised to the power of 11 over 10π‘₯ is equal to 𝑒 to the power of 11 over 10.

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