1. ## Evaluate the Limits

Question: If $f(9) = 9 , f'(9) = 4$ ,evaluate $\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$

2. Originally Posted by zorro
Question: If $f(9) = 9 , f'(9) = 4$ ,evaluate $\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$

As both the numerator and the denominator in your function approach 0 when x --> 9 and both are derivable functions around x = 9, you can Use L'Hospital's Rule here. The answer is 4.

Tonio

3. ## Is this correct?

Originally Posted by tonio
As both the numerator and the denominator in your function approach 0 when x --> 9 and both are derivable functions around x = 9, you can Use L'Hospital's Rule here. The answer is 4.

Tonio
Here is what i have done

$\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{ \sqrt{3} - 3}$

Using L'Hospitals rule

$g'(x)$ = $\frac{d}{dx} \left[ \sqrt{f(x)} - 3 \right]$ = $\frac{f'(x)}{2 \sqrt{f(x)}}$ ...............eq(1)

$h'(x)$ = $\frac{d}{dx} \left[ \sqrt{x} - 3 \right]$ = $\frac{1}{a \sqrt{x}}$ ..................eq(2)

Now taking lim of eq (1) and eq(2) we get

$\lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$

4. Originally Posted by zorro
Here is what i have done

$\lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{ \sqrt{3} - 3}$

Using L'Hospitals rule

$g'(x)$ = $\frac{d}{dx} \left[ \sqrt{f(x)} - 3 \right]$ = $\frac{f'(x)}{2 \sqrt{f(x)}}$ ...............eq(1)

$h'(x)$ = $\frac{d}{dx} \left[ \sqrt{x} - 3 \right]$ = $\frac{1}{a \sqrt{x}}$ ..................eq(2)

Now taking lim of eq (1) and eq(2) we get

$\lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$
Yes.