Find the limit. lim x → 0 x + 1 − 1 x \lim_{x\rarr0}\frac{\sqrt{x+1}-1}{x} lim x → 0 x x + 1 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> − 1

Here, we want to find the limit of the square root of x+oneoverx asxapproaches0. Now if I plugged 0 in for x here that would definitely make my denominator 0 and since I have radicals in this rational function I want to go ahead and multiply by the conjugate. Now here since I have my radical in the numerator I'm going to take the conjugate there of the square root of x+one-one by simply reversing reversing that sign in between my two terms. Remember, we don't need to reverse the sign that's actually under the radical, just the one that's in between my two terms. So the conjugate here is x+1outside of that radical having reversed that sign. Now since I'm multiplying by the new that by the numerator, I need to go ahead and multiply the denominator as well, the square root of x+1. Now, again, here, we wanna go ahead and fully multiply that numerator out in order to get stuff to cancel. So fully that numerator out will end up giving me x+1-one and then in my denominator I'm going to leave that as is with x times the square root of x+1. Now in my numerator here, this plus 1 minus 1, it cancels. And I am left to find the limit as x approaches 0 of x, that's all that's left in my numerator, over x times the square root of x+1. That is a mouthful. Now I can go ahead and cancel my x in my numerator with my x in the denominator. And I am finally left to find the limit as x approaches 0 of 1 over the square root of x plus 1 +1. Now, finally, here, I can go ahead and plug a 0 into my function to get my final answer. That'll give me 1 over the square root of 0+1. Now the square root of 0+1 is just going to be 1. So I am left here with 1 over 1+1 which gives me a final answer of 1 half as the limit of this function as x approaches 0. Now if you've made it to this point and you've been working through all of these practice problems, you have likely noticed that many of these practice problems that contain radicals have very specific formats. So be sure to recognize that and go ahead and work through some more problems because you'll be an expert in no time. Thanks for watching. Let me know if you have any questions here.

22. Limits & Continuity

Finding Limits Algebraically

Precalculus

NEW 22. Limits & Continuity

Finding Limits Algebraically

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Multiple Choice

Find the limit.
$\lim_{x\rarr0}\frac{\sqrt{x+1}-1}{x}$

$0$

$\frac12$

$1$

Does not exist

Verified step by step guidance

Step 1: Recognize that the problem involves finding the limit of a function as x approaches 0. The function given is $ \frac{\sqrt{x+1} - 1}{x} $.

Step 2: To solve this limit, consider using the technique of rationalizing the numerator. Multiply the numerator and the denominator by the conjugate of the numerator, which is $ \sqrt{x+1} + 1 $.

Step 3: After multiplying, the expression becomes $ \frac{(\sqrt{x+1} - 1)(\sqrt{x+1} + 1)}{x(\sqrt{x+1} + 1)} $. The numerator simplifies to $ (x+1) - 1 = x $.

Step 4: Substitute the simplified numerator back into the expression, resulting in $ \frac{x}{x(\sqrt{x+1} + 1)} $.

Step 5: Cancel the $ x $ in the numerator and denominator, leaving $ \frac{1}{\sqrt{x+1} + 1} $. Evaluate the limit as $ x \to 0 $, which simplifies to $ \frac{1}{2} $.