# Finding the equation of tangent line to the curve at the given point

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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Are you trying to find the tangent line or the slope of the tangent line? There is a big difference. The tangent line has an equation ($y = mx + b$ or something), and the slope of the tangent line is a number. – Jesse Madnick Sep 11 '11 at 17:52
Note that the equation of a line will have shape $ax+by=c$, where $a$, $b$, $c$ are constants. Anything that cannot be put into that shape cannot be right. – André Nicolas Sep 11 '11 at 17:57
"Find an equation of the tangent line to the curve at the given point." Not sure what the differences are in what you mentioned. – user138246 Sep 11 '11 at 17:58
@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 $\lim_{x\to 1}\frac{\sqrt{x} - 1}{x - 1}.$ – Jesse Madnick Sep 11 '11 at 22:46

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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Well in my book is explains to me to by using the equation (f(x)-f(a))/(x-a) I tried to do that and I get something that doesn't make sense. – user138246 Sep 11 '11 at 17:46
@Jordan: Note that basically $y = f(x)$ and $y_0 = f(x_0)$, so Jesse's answer says $m = (f(x) - f(x_0))/(x - x_0)$. Replacing $x_0$ with $a$ gives the same expression as your book. – TMM Sep 11 '11 at 17:49
I see. But are you sure that the formula $(f(x) - f(a))/(x-a)$ really gives you a formula for the tangent line? Because really, $(f(x) - f(a))/(x-a)$ gives the slope of the secant line between $(a, f(a))$ and $(x,f(x))$. On the other hand, the slope of the tangent line is $$\lim_{x\to a} \frac{f(x) - f(a)}{x-a}.$$ Note that neither of these formulas give an equation for a line, but rather a formula for computing the slope of a line. – Jesse Madnick Sep 11 '11 at 17:49
Yes it does, does that mean I make a = x? – user138246 Sep 11 '11 at 17:50
Neither Jesse's nor Austin's comments were rude or offensive. Do not flag them as such. – Zev Chonoles Sep 11 '11 at 18:30

• 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 $x$.