Tangent to the curve

Hello,

Quote:

Originally Posted by Lipticboven

I need help finding the tangent to the curve $y = \frac{3}{1 + \sqrt{x}}$ , at the point (4,1). Im really only having problems finding the gradent of the tangent, i already know the answer but everytime i try i either get the gradent as 0 or infinity which isnt correct.

Im not ment to be using differentiation, but instead using limits.

Let $f(x)=\frac{3}{1+\sqrt{x}}$

You have to find $f'(4)$ in order to find the gradient.

So since you have to use limits and not differentiation, remember that :
$f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$

So here :
$f'(4)=\lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$

$=\lim_{h \to 0} \frac{\frac{3}{1+\sqrt{4+h}}-\frac{3}{1+\sqrt{4}}}{h}$

$=\lim_{h \to 0} \frac{\frac{3}{1+\sqrt{4+h}}-1}{h} \cdot \frac{1+\sqrt{4+h}}{1+\sqrt{4+h}}$

$=\lim_{h \to 0} \frac{3-(1+\sqrt{4+h})}{h(1+\sqrt{4+h})}$

$=\lim_{h \to 0} \frac{2-\sqrt{4+h}}{h(1+\sqrt{4+h})} \cdot \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}$

(it's multiplying by the conjugate of the numerator and then use the identity $(a-b)(a+b)=a^2-b^2)$

$=\lim_{h \to 0} \frac{4-(4+h)}{h(1+\sqrt{4+h})(2+\sqrt{4+h})}$

$=\lim_{h \to 0} \frac{-h}{h(1+\sqrt{4+h})(2+\sqrt{4+h})}$

$=\lim_{h \to 0} \frac{-1}{(1+\sqrt{4+h})(2+\sqrt{4+h})}$

And now the limit is defined...