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in this problem, we want to show that the
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limit is X approaches for of the square root of
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X is equal to two. We want to do
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this using that slammed out to definition of a limit
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. So that means we want to show effort.
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Every Absalom greater than zero there is adult greater than
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zero such that thie, absolute value of the square
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root of X minus two is less than epsilon.
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Whenever the square root of X minus four is in
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between zero and delta time. So you know.
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So this here is the inequality that we want to
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satisfy and we want to find tell to that would
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always go always work the matter which positive number Absalon
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ticks on. So this here is true then,
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yeah, we can take both sides of this equation
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and multiply them by the conjugation of thie absolute value
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of the square root of X minus two. So
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we kind of multiply besides by the conscript of it's
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great of X minus two and noticed that the Contra
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kit is in absolute value. So it's always positive
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, so we don't have to worry about flipping this
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inequality sign or anything like that. So since that's
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true and we combine, these terms will see that
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they actually equal. He have some value of X
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minus four. Yeah, So that tells us that
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that is less than Absalon Times, the Contra kit
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. So now we have this expression here in terms
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of absolute and X two variables and three what we
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wanted. Teo is bound one of them, so
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we're going to bound X so that it's somewhat fixed
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. So let's let X be in the interval one
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tonight so that it must be the case that the
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square root of X is in the interval. 123
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and the square root of X close to is in
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the interval 3 to 5. Because square root of
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X is just shifted over on the number line by
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two silly kind of add to to the end points
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. And here we kind of take discovered of the
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end points. So an interval notation. This says
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that three is less than this query to vex close
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to when that's less than five. There's also implies
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by taking reciprocal is that 1/3 is greater than one
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over the square root of X plus two when that's
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creator and one over five. So this will be
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very important to us in a few steps. So
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no, let's let UPS alone be given. And
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let's let Delta equal to three times up Salon.
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Then, if we have that, zero is less
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than that's the value of X minus four, which
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is always true since the absolutely anything should be greater
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than zero. And that's less than Delta, which
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is three times up. Salon. It must be
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the case that the factory ization of X minus four
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, which we saw right here, is also in
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that interval because they're equal, sir. So we've
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kind of just written this absolute value in its factored
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form. So this implies that this value right here
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it's less than three absalon divided by this conjugal value
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here, if we kind of divided over on both
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sides, therefore we have that absolute value of the
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square root of X minus two is less than three
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Absalon divided by this value. Remember, we have
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at this is bounded by 1/3 has bounded above by
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1/3. So that's less than three Absalon times 1/3
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which, of course, is less than absolute when
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we're done. So you picked experience for once.
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Nine. If Absalon, we've chosen to be particularly
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large, we might run into some issues, but
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we can always adjust and pick a small enough delta
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to match the up salon because of this relationship here
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.