find-lim-_x-rightarrow-0-fracxln-x1-2

Name: find lim _x rightarrow 0 fracxln x1 2
Uploaded: 2019-09-19
Duration: 1 min 15 s
Channel: Numerade
Description: Find $\lim _{x \rightarrow 0^{+}} \frac{x}{\ln (x+1)}$

Question

Find $\lim _{x \rightarrow 0^{+}} \frac{x}{\ln (x+1)}$

Taylor Shimono · Accepted Answer

So for this problem, we&#x27;re finding the limit as X approaches zero from the right hand side of X over the natural log of X plus one. And we know that our first step in finding a limit is just going to be plugging in whatever X is approaching. So when you plug this in here, we get zero over the natural log of one which is going to end up being zero over zero. So since this is in indeterminate form, we&#x27;re going to go ahead and apply Loki Toll troll and that tells us that our limit with the same bounce so it&#x27;ll still be as experts zero from the right hand side of the derivative of our new writer over the derivative of already nominator is going to be equal to the same thing. So when we take our derivatives, we get the limit as experts zero from the right hand side of the derivative of X, just one over the derivative of the natural log of X plus one is one over X plus one. And so, since we&#x27;re dividing by a fraction and our denominator, this is just going to become the limit as expert zero with the right hands from the right hand side of X plus one, and from here we can see that we&#x27;re just going to be able to plug in zero for X and evaluate our limit to be equal to.

Ahmed Ibrahim · Answer

this question. We can take FF X equal tow X our negative one overland X So if we took Len eggs, we can get negative then X over next, which is equal to negative one. Therefore, if we took lam it when X approaches to issue for our farms and who fix, we can write it as e or Glenn If we fix and lend FX. We just told with as negative one. So it would be people negative corn, which is equal or former solution e one over Thanks.

Ma. Theresa Alin · Answer

to evaluate the limits As X approaches zero from the right of one over x minus Ln of X. We first write this into The limit as X approaches zero from the right of one over X the limit. As X approaches zero from the right of Helena vex. Now, for the first limit, If X approaches zero from the right then we know the X value there is a small positive number. And so the value of one over X must be approaching positive infinity. That means for the first time we have positive infinity minus for the limit of L innovex. As X approaches zero from the right, we recall the graph of al innovex and here we see that as we move closer to zero, the value of Ln of X approaches negative infinity. And so and here we have negative infinity and so infinity minus negative infinity, that&#x27;s plus infinity. Or this is going to be positive infinity. And so this is the value of the limits.

Zachary Mitchell · Answer

here we have the limits as X goes to zero from the positive side of one minus natural Log X divided by E to the one over X. Now, if we try to do this limit notice in the numerator, we&#x27;re going to get a positive infinity because the natural log by itself as a tends to zero from the right hand side equals negative infinity. So the numerator equals positive affinity the denominator We get a positive infinity. So we&#x27;ve infinity over infinity. That&#x27;s an indeterminant form and we can go ahead and use low Patel&#x27;s rule. We&#x27;re gonna write in hell h of this equal sign indicating that that&#x27;s what we do for this next step. Okay, so the derivative of the numerator is negative. One over X, the derivative the denominator. We&#x27;re gonna have to do a chain rule. Let&#x27;s see, So we get a negative X to the negative, too. Then we saw this and then this next step is simplifying. It simplifies here we get okay. And now this is a limit that weaken do because of the the numerator goes to zero. The denominator goes to infinity and that&#x27;s it

Zachary Mitchell · Answer

We&#x27;re starting with the limit as X goes to zero from the positive side of negative natural log X rays to the X. Let&#x27;s go ahead and assign that to variable L And then we&#x27;re gonna take the natural log of each side. And now, for our next step, we&#x27;re going to switch the natural log and the limit. And we can do this because natural August continuous. Okay? And so let&#x27;s go ahead and rewrite this thing as a fraction using the natural log properties so you can take that X and bring it down. Rewrite it as one over X. Okay, um, now, notice that the numerator does in fact make sense because, as zero has extends to zero from the positive side, natural law give extends to negative infinity. So we have the numerator attending to positive infinity on the top and the denominator tending to positivity on the bottom. So we&#x27;re good to go to use Lou petals rule. Let&#x27;s go ahead and do that one over a natural log X times negative one over X and then the denominator. We have negative one over X squared, and now we can simplify this. Um, and this works out to be negative. X over natural log X and this we can evaluate it. Zero Okay, so we have natural law. Go back and we have natural log of l is equal to zero. That implies that are limits. L is equal to one and we&#x27;re done.

Zachary Mitchell · Answer

So we&#x27;re looking at the limit as X goes to it. Positive infinity of natural log of X raised to the one over X. Let&#x27;s set that equal toe l Now let&#x27;s take the limit. Excuse me. The natural log of each side. Okay. And our next step is to interchange the limit with the natural log. And we can do this because natural log. It&#x27;s continuous. Okay, let&#x27;s go ahead and use our log property to bring that one that one over X down. So now we have natural log of natural. Lagerback&#x27;s divided by X. And if we try to evaluate this limit, we get infinity over infinity. So we&#x27;re set up to use low petals rule. Let&#x27;s go ahead and do that. So the derivative, the top is one over. Natural log X times one over X by chain room to remove. The bottom is just one. Okay. And this limit is something that we can easily evaluate. Zero. Okay, so we have that The natural log of LZ zero. This implies that l is equal to one. That&#x27;s it. We&#x27;re done

Problem

Jacob is now 12 years younger than Michael. If 9 …

Problem 10 Medium Difficulty

Find $\lim _{x \rightarrow 0^{+}} \frac{x}{\ln (x+1)}$

Answer

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