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I am trying to find the tangent line at $y=\sqrt{x} $ , (1,1)

I know that I need to use the tangent line equation and I end up with $(\sqrt{x} - 1)/1-1$

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    @Jordan: I believe André is saying that there is$a$difference between "an _equation_ of the tangent line to a curve" and the _number_ given by $\lim_{x\to 1}\frac{\sqrt{x} - 1}{x - 1}.$2011-09-11

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EDIT: Let's clarify a couple of things.

The slope of the secant line between $(a, f(a))$ and $(x,f(x)))$ is $\frac{f(x) - f(a)}{x-a}.$

The slope of the tangent line at $(a, f(a))$ is $\lim_{x\to a}\frac{f(x) - f(a)}{x-a}.$

To find the equation of a tangent line, one needs to use the point-slope formula, which I've explained below.


Now, in your case, $f(x) = \sqrt{x}$, and we have $a = 1$, $f(a) = 1$. So the slope of the tangent line is $\lim_{x\to 1}\frac{\sqrt{x} - 1}{x-1}.$ Now we have to evaluate this limit.

If we try to evaluate this limit by just plugging in $x = 1$, we get $0/0$, which is a problem (dividing by zero is bad), so we need a new strategy.

Idea: When evaluating the limits of fractions, a good trick is to multiply the top and bottom by the "radical conjugate." So:

$\begin{align} \frac{\sqrt{x} - 1}{x-1} & = \frac{\sqrt{x} - 1}{x-1}\frac{\sqrt{x} + 1}{\sqrt{x} + 1} \\ & = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x-1)(\sqrt{x} + 1)} \\ & = \frac{x - 1}{(x-1)(\sqrt{x} + 1)} \\ & = \frac{1}{\sqrt{x} + 1}. \end{align}$

Now we can evaluate $\lim_{x \to 1} \frac{\sqrt{x} - 1}{x-1} = \lim_{x\to 1}\frac{1}{\sqrt{x} + 1}$ by plugging in $x = 1$ no problem. This will give us the slope of the tangent line. If you want the equation of the tangent line, you need the point-slope formula, explained below.


The point-slope formula says that a line with slope $m$ that passes through $(x_0, y_0)$ has an equation of the form $y - y_0 = m(x-x_0).$

In your case, the tangent line passes through $(1,1)$, so you can plug in $x_0 = 1$, $y_0 = 1$. We'll also have the slope, $m$, from the previous section once we evaluate that limit (which I leave to you to do).

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    @Jordan: If the point in question is $(1,1)$, then this tells us that $a = 1$ and $f(a) = 1$. We don't need to substitute anything in for $x$ or $\sqrt{x}$ since otherwise we could not take the limit as $x \to 1$. And again: after we take the limit, we should get$a$_number_, not a function nor an equation. To get the _equation_ of the line, we need to use the point-slope formula.2011-09-11
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Several comments:

  • You shouldn't write $(\sqrt{x} - 1)/1-1$ if you mean $(\sqrt{x} - 1)/(1-1)$; remember the conventions on order of operations.
  • If you put $1$ in place of $x$ in $(\sqrt{x} - 1)/(x-1)$, what you get is $(\sqrt{1} - 1)/(1-1)$. This is $0/0$.
  • The expression $(\sqrt{x} - 1)/(x-1)$ gives the slope of a secant line, not of a tangent line.
  • Since $0/0$ is undefined, in order to find the slope of the tangent line, you need to find $\lim\limits_{x\to1} (\sqrt{x} - 1)/(x-1)$, rather than simply plugging in $1$ in place of $x$.
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    My biggest problem is I don't know what f(x) is and what f(a) is, is is y and x or what? My book does not explain this anywhere or even show work for anything really, they just jump to the answer. Every example the book uses a = 1 but the points are always 1,1. Whoever made this book is an idiot, there is no definition of 1 in this book. Is is either x or y.2011-09-11
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y=root of x, or simply f(x)= root of x at the point (1,1) by deriving the root of x we get x^1/2 which is equal to 1/2.root of x plug in the x value from (1,1) and get 1/2 we know y,m,and x so we only need to find c in this equation, y=mx+c then we get 1=1/2*1+c c=1/2 so our equation will be y=1/2x+1/2