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# (a) Estimate the value of the limit $\displaystyle \lim_{x \to 0}(1 + x)^{1/x}$ to five decimal places. Does this number look familiar?(b) Illustrate part (a) by graphing the function $y = (1 + x)^{1/x}$.

## (a) Let $h(x)=(1+x)^{1 / x}$$\begin{array}{|c|c|}\hline x & h(x) \\\hline-0.001 & 2.71964 \\-0.0001 & 2.71842 \\-0.00001 & 2.71830 \\-0.000001 & 2.71828 \\0.000001 & 2.71828 \\0.00001 & 2.71827 \\0.0001 & 2.71815 \\0.001 & 2.71692 \\\hline\end{array}$$It appears that$\lim _{x \rightarrow 0}(1+x)^{1 / x} \approx 2.71828,$which is approximately eIn Section 3.6 we will see that the value of the limit is exactly$e\$

Limits

Derivatives

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

and this is problem boarding. Seven of the Stuart Calculus eighth edition Section two point two Partying. It's the estimate, the value of the limit limit as experts zero of the corn eighty one plus x razed to the one rx power two five decimal places. Does this number look familiar? Well, it's first estimate, the value this limit. And as we know, as we approach zero, we want to choose values. To prevent an estimation, we'LL choose values very close to zero both from the left and from Lori and to double check. We can't directly plug in zero here because Syria would be in the denominator. Therefore, we know that for this function, the value at X equals zero is undefined. Two. We pull up a table and we choose values from the right and from the left that approach zero closer and closer and we take the function one plus x rays to the one heretics power and we see how the function changes. The closer we get to zero again. Confirming that X equals zero is an undefined point for the function. As we get closer and closer to zero from the right and from the left, we approached the same value here as you can see. Two five decimal places we approached value two point seven one eight two eight and we're on the right tattoo right here for a party. Two point seven one hit two eight and party asks. This number looks familiar. Just under is a very close approximation. Two. The mathematical constant e And that is the solution from hard, eh, Here then he's not exactly two point seven, twenty to eight. But it is exactly the solution to its limit because as we get closer and closer to X equals zero disfunction gets closer and closer to the number e. We're going to illustrate this limit imparting by graphing the function Michael's good. The quantity one plus six race to the one over X power toe. We take the values, we plot the function and we see as the approach X equals zero from the left and from the right. We're approaching that value. Ah, that we just confirmed about you. E, which is approximately two point seven one eight two eight

#### Topics

Limits

Derivatives

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp