Determine the following limits.
lim h→0 √5x + 5h − √5x ...

Question

Determine the following limits.

lim h→0 √5x + 5h − √5x / h, where x>0 Is constant

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with limits. This problem says, determine the limit as N approaches zero or the expression which is the quantity of the square root of the quantity of nine W plus nine nn quantity minus the square root of nine WN quantity divided by N where W is greater than zero is a constant. We're given four possible choices. As our answers. For choice A, we have one divided by the quantity of three multiplied by the square root of W. For choice B, we have three divided by the quantity of two multiplied by the square root of W. For choice C, we have one divided by the square root of W. For choice D, we have nine divided by the square root of W. So in this problem, we're asked to find the limit as N approaches zero of the expression, which is the quantity of the square root of the quantity of nine W plus nine nn quantity minus the square root of nine W in quantity divided by N. Now, if we just set N equal to zero and try to evaluate this limit that way. We'll actually get zero divided by zero, which is an indeterminate form. So to actually find the limit, we're going to multiply our expression by one in the form of the quantity of the square root of the quantity of nine W plus nine N in quantity plus the square root of nine W in quantity divided by the quantity of the square root of the quantity of nine W plus nine N in quantity plus square root of nine W. So when we multiply our numerator and our denominator out, what we're going to get is the limit as N approaches zero of the quantity of nine W plus nine N minus nine W. And that quantity is divided by the quantity of N multiplying the quantity of square root of the quantity of nine W plus nine N in quantity plus the square root of nine W. So here we need to simplify our numerator. So we'll have nine W plus nine N minus nine W. So the nine wws will cancel out since we have one positive quantity and one negative quantity. And then this is going to leave me with the limit as N approaches zero of nine N divided by the quantity of N multiplying the quantity of the square root of the quantity of nine W plus nine in in quantity plus square root of nine W. Now the um factor of N will cancel out since I have an N in the numerator and into the, in the denominator. And at this point, I can now substitute NN equal to zero into our expression. So doing that, we will get this equal to nine divided by the quantity of the square root of nine W plus the square root of NW. We can simplify this expression a bit. So this is going to be equal to nine divided by the quantity of two multiplied by the square root of nine W. Now the square root of nine is three and three times two is six. So we can simplify this even more. So this is gonna be equal to nine divided by the quantity of six multiplied by the square root of W. And we can divide the numerator and denominator by three to simplify our fraction even more. So this is going to be equal to three divided by the quantity of two multiplied by the square root of W. So this is the answer that I'm looking for. And when I compare the answer that we calculated to the answer choices, the correct answer choice turns out to be answer choice B so I hope that this explanation has been useful and I'll see you in the next video.

Determine the following limits.
lim h→0 √5x + 5h − √5x / h, where x>0 Is constant

Key Concepts

Limits

Derivative

Rationalization

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