Question Video: Solving Limits by Using the Definition of Euler’s number | Nagwa Question Video: Solving Limits by Using the Definition of Euler’s number | Nagwa

Question Video: Solving Limits by Using the Definition of Euler’s number Mathematics

Determine lim_(π‘₯ β†’ ∞) (1 + (1/π‘₯))^(4π‘₯).

02:47

Video Transcript

Determine the limit as π‘₯ approaches ∞ of one plus one over π‘₯ all raised to the power of four π‘₯.

We’re given the limit which we need to evaluate. And the first thing we can try and do is just evaluate this limit directly. We see that our limit is as π‘₯ is approaching ∞. And we know as π‘₯ approaches ∞, the reciprocal function one over π‘₯ approaches zero. So, the inner part of our parentheses are approaching zero. However, as π‘₯ approaches ∞, four π‘₯ is approaching ∞. So, trying to evaluate this limit directly, we get one to the power of ∞, and this is an indeterminate form. So, we’re going to need to try some other method of evaluating this limit. Instead, we need to notice the limit given to us in the question is very similar to the limit in the definition of Euler’s number. So, we need to recall the following limit result.

We know the limit as π‘₯ approaches ∞ of one plus one over π‘₯ all raised to the power of π‘₯ is equal to Euler’s constant 𝑒, and it’s worth committing this limit to memory. And it’s also worth pointing out sometimes you’ll see this written in terms of the variable 𝑛. However, we’re given the limit in terms of π‘₯, so we’ve just rewritten this in terms of π‘₯. And we can see the limit given to us in the question is almost exactly in this form. However, in our exponent, instead of just having π‘₯, we have four π‘₯. So, we want to rewrite this limit into a form where we can just use our limit result. And to do this, we’re going to need to use our laws of exponents.

First, we need to recall 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. And we want to use this to rewrite our limit into a form where we can use our limit result. This means we’re first going to want to rearrange the two factors in our exponent. By doing this and then using π‘Ž is equal to one plus one over π‘₯, 𝑏 is equal to π‘₯, and 𝑐 is equal to four, we’ve rewritten our limit as the limit as π‘₯ approaches ∞ of one plus one over π‘₯ all raised to the power of π‘₯ all raised to the power of four.

But we can’t yet use our limit result because we’re raising this all to the fourth power. So, we’re going to want to apply the power rule for limits. One version of this tells us for any integer 𝑛 and real constant π‘Ž such that the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ exists, then 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 to the 𝑛th power. And it’s worth pointing out this is also true if we’re taking limits at ∞. We want to apply this with 𝑛 set to be four, π‘Ž equal to ∞, and 𝑓 of π‘₯ to be one plus one over π‘₯ all raised to the power of π‘₯.

And It’s worth pointing out we know the limit as π‘₯ approaches ∞ of 𝑓 of π‘₯ exists because it’s equal to 𝑒. So, by the power rule for limits, we can rewrite our limit as the limit as π‘₯ approaching ∞ of one plus one over π‘₯ all raised to the power of π‘₯, and then we raise this limit result to the power of four.

And now, we can just evaluate our inner limit. We know what it’s equal to; it’s equal to Euler’s constant 𝑒. So, by replacing this limit with 𝑒, we get 𝑒 to the fourth power, which is our final answer. Therefore, we were able to show the limit as π‘₯ approaches ∞ of one plus one over π‘₯ all raised to the power of four π‘₯ is just equal to 𝑒 to the fourth power.

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