Using limits to find an equation of a tangent line to the graph of f at given point

**AMaccy** · May 31st 2009, 11:39 AM

Given:

$\sqrt{x+2}$ at the point $(7,3)$

I need to use the limit definition of :

$m= \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)- f(x)}{\Delta x}$

to find an equation of the tangent line to the graph of $f$ at the given point (above).

**skeeter** · May 31st 2009, 12:31 PM

Originally Posted by AMaccy

Given:

$\sqrt{x+2}$ at the point $(7,3)$

I need to use the limit definition of :

$m= \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)- f(x)}{\Delta x}$

to find an equation of the tangent line to the graph of $f$ at the given point (above).

let $h = \Delta x$

$f(x) = \sqrt{x+2}$

$f(x+h) = \sqrt{x+h+2}$

$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$\lim{h \to 0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}$

$\lim{h \to 0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{ x+2}}$

$\lim{h \to 0} \frac{(x+h+2)-(x+2)}{h(\sqrt{x+h+2}+\sqrt{x+2})}$

$\lim_{h \to 0} \frac{1}{\sqrt{x+h+2}+\sqrt{x+2}} = \, ?$

sub in $7$ for x as your last step after you find $f'(x)$

find the tangent line equation using the point-slope form of a linear equation

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