Find the limit. lim x → 3 x − 3 x − 3 \lim_{x\rarr3}\frac{\sqrt{x}-\sqrt3}{x-3} lim x → 3 x − 3 x <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> − 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

In this problem, we want to find the limit of the square root of x minus the square root of 3 over x minus 3 as x approaches 3. Now here if I were to plug that 3 in directly my denominator would be 0 which I do not want to happen and since I have radicals in this rational multiplying by the conjugate. Now here we're going to take the conjugate of our numerator because that's where our radicals are. So multiplying by the conjugate here the square root of x plus the square root of 3 just reversing that sign in between my two terms. Now since I'm multiplying the numerator I of course need to multiply the denominator by the same thing the square root of x plus the square root of 3 and we can go ahead and fully expand that numerator out multiplying it all together. So I'm still taking the limit as x approaches 3, but fully multiplying that numerator out will end up giving me x minus 3. Now in my denominator, I am not going to fully multiply that out because we'll see what cancels out and I'm going to leave that as x minus 3 times the square root of x plus the square root of 3. Now here I have that common factor x minus 3 that cancels out. And then I am left to find the limit of 1 over the square root of x plus the square root of 3 as x approaches 3 because that's all I'm left with here. So now I can go ahead and plug that 3 in to get my final answer. So I have one over the square root of 3 plus the square root of 3 and combining those square roots on the bottom this ends up being 1 over 2 root 3. But we don't wanna have that radical in the denominator. We wanna go ahead and rationalize here by multiplying by the square root of 3 over the square root of 3. Now in doing that, that ends up getting me the square root of 3 over 6 with that denominator rationalized. So this is my final answer here for the limit of this function as x approaches 3. Let me know if you have any questions here, and I'll see you in the next one.

1. Limits and Continuity

Finding Limits Algebraically

Calculus

1. Limits and Continuity

Finding Limits Algebraically

Calculus 때문에 힘들어하고 계신가요?

저희를 신뢰하는 수천 명의 학생들이 시험 합류하세요!

객관식

Find the limit.
$$\lim$_{x$\rarr$3}$\frac{\sqrt{x}$-$\sqrt$3}{x-3}$

$2$\sqrt$3$

$0$

$\frac{\sqrt3}{6}$

Does not exist

검증된 단계별 안내

First, recognize that the given limit is in the indeterminate form 0/0 as x approaches 3. This suggests that we need to simplify the expression to evaluate the limit.

To simplify, multiply the numerator and the denominator by the conjugate of the numerator, which is $ \sqrt{x} + \sqrt{3} $. This technique helps eliminate the square roots in the numerator.

After multiplying, the numerator becomes $ (\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3}) = x - 3 $, and the denominator becomes $ (x - 3)(\sqrt{x} + \sqrt{3}) $.

Now, the expression simplifies to $ \frac{x - 3}{(x - 3)(\sqrt{x} + \sqrt{3})} $. Cancel the common factor $ x - 3 $ from the numerator and the denominator.

The simplified expression is $ \frac{1}{\sqrt{x} + \sqrt{3}} $. Now, substitute $ x = 3 $ into this expression to find the limit.

6:13m

다음 시청

마스터 Limits of Rational Functions with Radicals가 Nick Kaneko 짧은 영상으로 설명해 줍니다

Finding Limits Algebraically 연습 세트

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전문가 영상 설명

Calculus 때문에 힘들어하고 계신가요?

Find the limit.limx→3x−3x−3\(\lim\)_{x\(\rarr\)3}\(\frac{\sqrt{x}\)-\(\sqrt\)3}{x-3}limx→3x−3x−3

다음 시청

Finding Limits Algebraically 연습 세트

Find the limit.
$\(\lim\)_{x\(\rarr\)3}\(\frac{\sqrt{x}\)-\(\sqrt\)3}{x-3}$