evaluate-lim-_x-rightarrow-16-fracsqrt4x-2x-16

Name: evaluate lim _x rightarrow 16 fracsqrt4x 2x 16
Uploaded: 2020-06-20
Duration: 2 min 39 s
Channel: Numerade
Description: Evaluate $\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}$.

Question

Evaluate $\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}$.

Carter Rogers · Accepted Answer

here. We&#x27;re going to have to get a little tricky with the way that we simplify this function. So we&#x27;re looking at the limit as Ex Purchase 16 of the fourth root of X minus two, divided by X minus 16. So, normally we would rationalize the numerator or wherever a radical is so we&#x27;ll go ahead and do that. Well, multiply this by the fourth root of X and then a plus two. And of course, we have to do the same to the denominator. Go ahead and do that. This equals the limit as X approaches 16 432 Vex times 1/4 root of X is the square root of X. These terms should cancel. And then we&#x27;re left with a minus four because positive, two times negative, too. And then the denominator gets a little messy and we&#x27;ll go ahead and leave it as it is. We won&#x27;t multiply that out. We have X minus 16 times 1/4 root of X plus two. And then again, we still have this radical here. So what we&#x27;re gonna do is we&#x27;re actually going to rationalize it one more time. So multiplying by the square root of X Plus four and again do the same to the denominator. We&#x27;ll go down here and we&#x27;ll say the limit as exit purchase 16. The numerator becomes brood X times your index, which is just X minus four times plus four, gives us minus 16 and then the denominator gets a little messy. But again, we won&#x27;t multiply it out. X minus 16 before through decks plus two that stays. And then we&#x27;re also multiplying by the square root of X plus four. So that&#x27;s our entire denominator. And then finally we can cancel, and then we&#x27;re left with the limit as exit purchase 16 of one over the four through of X Plus two. We&#x27;re multiplying that by the square root of X Plus four, and finally, we can go ahead and plug in. Our value will go ahead and do that. We have one over the four through of 16 which is to plus two other parentheses. E. We have the square root of 16 which is four, so we have four plus four. This gives us 1/2 plus two is four. Four plus four is eight, so four times eight, which is 32 has your final answer

Arin Asawa · Answer

So we&#x27;re given the limit as eggs approaches 16 off square root of X minus 16. And so let&#x27;s evaluate it. Well squared of X minus 16 is going to be defined for all eggs that are, um, more than or equal to zero. So this function right over here is gonna have a domain of X and more than or equal to zero. Because if you have negative than you&#x27;re going into imaginary numbers and we don&#x27;t want that as old, since it&#x27;s defined for all adds more than or equal to zero are we know it&#x27;s continuous on X is on this interval. Over here are functions continuous? Um, we know that it&#x27;s continuous. Add X equal 16. Uh, so we know that squared of X minus 16 is continuous. Add ex Eagle 16 and if it&#x27;s continuous there than the limit, then this limit over years just gonna be the same thing as the value of the function at the X value, because again, are functions continuous at X equal 16 so we can just plug or that into our function. And so this limit over here is going to be equal to the square root of 16 minus 16. And so that&#x27;s going to be school. The squared of 16 is four for minus, um, four. Minus 16 is going to be negative 12. And so that right over there is our limits.

Cinsy Krehbiel · Answer

what happens to this function? F as Ex gets very large in the negative direction. In order to evaluate that, we want to factor out this squared X squared. The spirit of X squared is equal to the absolute value of x x, less than zero. The absolute value of X equals negative X right. If X is negative one, the absolute value of that is negative. Negative one, and it will be the same for all negative axes. Factoring that out, it&#x27;s weird of X squared leaves under the radical here, a 16 plus one over X that&#x27;s our numerator will leave the denominator alone. I just figured out a moment to go that the square root of X squared is equal to negative x negative X times the square root of 16 plus one over X as X approaches infinity. Whether it&#x27;s negative or positive, we end up with the fraction that has a very large denominator in either the negative or positive direction. In a very, very large denominator in a fraction, the larger the denominator, the closer let&#x27;s say approaches zero that fraction becomes traction, approaches zero, and therefore it will contribute very, very little to the value of f of X as X approaches infinity. Whether it&#x27;s cluster negative for very, very large exes, F of X is approximately equal to the square root of 16. Very, very close. It&#x27;s actually negative 16. Even though these X is canceled, this negative one is still there. So the limit as X approaches negative infinity of our function F of X is equal to negative four.

Yaw Asomani · Answer

limits. Execution 16 off. Four minus Screwed off that 16 x minus X squared. So when you put c seen there, this is gonna be 16 squared. Mind assisting square, which is zero. Right. And when you put 16 here at the numerator, a swell is gonna be screwed of 16 which is four for minus forces of zeros. A little bit 00 which is indeterminate. Right, So we have to do more work. And what we&#x27;re gonna do is, you know, the contribution thing gonna multiply about a conjugation. So this is the conjugation off that right? So this is gonna be the first term squared, and mine is a second term squared, right? We&#x27;ve been doing this throughout, and then this is gonna be 16 x minus X squared and then four plus that right? And ain&#x27;t taking the limits. X approaches 16 of that. Now I&#x27;m gonna do a bit off the appellations, right. I&#x27;m gonna factor this one, see excess comments. I&#x27;m gonna factor it out. So when I found her it out, I&#x27;m gonna have 16 minus x left, and then here is gonna be this, right? And then this is canceling this still have limited a problem. Teoh, this four, right? I&#x27;ve just reduce it to this form, which makes my life a whole lot easier. A lot better. Right now. When I put 16 there, that is gonna be 1/16. Four plus put. 16 here. Screwed of 16 is four. Right? So four plus four. So when you do this relationship, Lee put it on the calculator. You can get one over 1 to 8. Great. And that becomes your answer.

Amy Jiang · Answer

So if you knows here we plug in using Drexel, we get division by zero. So instead of doing that, let&#x27;s multiply by our conjugal base, which is four plus square root of that. And we&#x27;re multiply and divide by that. Okay, so we get the limit as expert to 16 we will keep our did not are numerator as is and then we&#x27;ll multiply in our, um are deny me here. So we get 16 minus I and I&#x27;ll let&#x27;s fact throughout the next. So we get negative negative 16 plus X like so And then we can Catto are X minus 16. I know that X cannot be equal to 16. OK, and now let&#x27;s use drugs up. So we get, um we have the negative of for plus the square root of 16. And that&#x27;s equal to negative for us, for which is equal to and they&#x27;re gonna eight

Jason Orozco · Answer

I want to find the limit as X goes to zero of the function X divided by Route 16 X plus one minus one all to the power of 1/3. Um, so first will just use the limit rule for roots up. Teoh Rewrite. This adds parentheses limit as X goes to Ciro X Group 16 X plus one. But it&#x27;s one, and we&#x27;re raising the entire limit to the 1/3 power. So that allows us to sort of disregard that, uh, 1/3 exponents until after we brought you the inner limit. Uh, so to do that, since we have Well, first we want toe check. Direct substitution. We can see that if we plug in zero into the denominator, we&#x27;re gonna get route 16 times their A plus one minus one. Uh, which is route one minus one, which is zero. So direct substitution is not going to work. Um, you to try something else? So since we have a radical, uh, minus one, we&#x27;re going Teoh, let&#x27;s multiply at the top of the bottom by the contra kit. So that&#x27;s going to give us get the parentheses going, uh, limit as X goes to zero of X times and then the contra git is route 16 x plus one plus one changed sign divided by, um, and then multiplying that that, uh, the same expression by the original denominator, the radicals gonna cancel out. So we&#x27;ll get 16 x plus one, and then the minus one times plus one will give us minus one so the ones will cancel out. So the denominator is just 16 X. Uh, so this expression is still equal to our original limit. And now this is a zero, Um, and so we can see that we tried direct substitution here plugging zero and gives us zero in the denominator. So that&#x27;s not gonna work. Um, but we do have an X, and he has a factor of the numerator and the denominator. So if we just look at the left or right limit by itself. So let&#x27;s say we just look at the limit as X goes to zero plus well, on the right, X is not equal to zero. Which means that if we wanted to find the right limit of this expression, we could weaken, cancel out the exes from the numerator and the denominator and get six, Route 16 x plus one plus one divided by 16. Um, and so this is this no longer has a serum denominator because he has 16 so we&#x27;ll just, uh, used directs. Institution plug zero into X 16 times zero plus one plus one worker 16. Um, and that gives us to over 16 war 1/8. Um, so we can see that if we were to take clear to change this to the limit as excuse to zero minus, we could apply the exact same calculations. Excellent. Cancel out, because X isn&#x27;t zero to the left of zero. Um, and then we would use direct substitution. So the limit as X goes to zero minus is also 18 since the left and right limits are equal. Um, we can conclude that the original limit as X goes to zero up here is also equal to one. It just by definition, um, so, uh, continuing from up here, we can plug in 1/8 for the limit, and it&#x27;s still raised to a 1/3 power, and that gives us 1/2. So our limit is excuse to zero is just 1/2

Problem

Creating functions satisfying given limit conditi…

Problem 100 Hard Difficulty

Evaluate $\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}$.

Answer

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