If #f(9)=9# and #f^'(9)=4#, then what is the value of #lim_(x to9)(sqrt(f(x))-3)/(sqrtx -3)?#

Question

Evelyn Agard · Accepted Answer

#lim_(xto9) (sqrtf(x)-3)/(sqrtx-3)=4#

We know that,

#(1)f'(c)=lim_(xtoc) (f(x)-f(c))/(x-c) ,# take #x=9#

#=>f'(9)=lim_(xto9) (f(x)-f(9))/(x-9) ,where, f(9)=9#

#=>f'(9)=lim_(xto9)(f(x)-9)/(x-9) ,where, f'(9)=4#

#=>color(brown)(lim_(xto9)(f(x)-9)/(x-9)=4.....(2)#

We take,

#L=lim_(xto9) (sqrtf(x)-3)/(sqrtx-3)# #color(white)(L)=lim_(xto9)(sqrtf(x)-3)/(sqrtx-3)xxcolor(red) ((1))xxcolor(blue)((1))#

#color(white)(L)=lim_(xto9)(sqrtf(x)-3)/(sqrtx-3)xxcolor(red)((sqrtf(x)+3)/(sqrtf(x)+3))xxcolor(blue)((sqrtx+3)/(sqrtx+3))#

#color(white)(L)=lim_(xto9)((sqrtf(x))^2-(3)^2)/((sqrtx)^2-(3)^2)xx(color(blue)(sqrtx+3))/(color(red)(sqrtf(x)+3))#

#color(white)(L)=color(brown)(lim_(xto9)(f(x)-9)/(x-9))xxlim_(xto9)(sqrtx+3)/(sqrtf(x)+3)...tocolor(brown)([use (2)]#

#color(white)(L)=color(brown)(4)xx(sqrt9+3)/(sqrtf(9)+3)#

#color(white)(L)=4[(3+3)/(sqrt9+3)]#

#:.L=4{6/6}=4#

Paisley Johnson · Answer

undefined To find the value of $ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} $, we can use L'Hôpital's Rule since both the numerator and denominator approach 0 as $ x $ approaches 9.

First, let's find $ f'(x) $ using the given information:

$$ f'(9) = 4 $$

Next, we'll apply L'Hôpital's Rule to the expression:

$$ \lim_{x \to 9} \frac{\frac{1}{2\sqrt{f(x)}} \cdot f'(x)}{\frac{1}{2\sqrt{x}}} $$

Now, substitute $ x = 9 $ and $ f(9) = 9 $:

$$ = \frac{\frac{1}{2\sqrt{9}} \cdot 4}{\frac{1}{2\sqrt{9}}} $$

$$ = \frac{\frac{4}{6}}{\frac{1}{6}} $$

$$ = \frac{4}{1} $$

$$ = 4 $$

So, the value of $ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} $ is 4.