# Evaluate the Limits

• Oct 11th 2009, 10:59 PM
zorro
Evaluate the Limits
Question: If $\displaystyle f(9) = 9 , f'(9) = 4$ ,evaluate $\displaystyle \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3 }$
• Oct 12th 2009, 02:17 AM
tonio
Quote:

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

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
• Dec 8th 2009, 03:39 PM
zorro
Is this correct?
Quote:

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

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

Using L'Hospitals rule

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

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

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

$\displaystyle \lim_{x \to 9} \left[ \frac{g'(x)}{h'(x)} \right]$ = $\displaystyle \lim_{x \to 9} \left[ \frac{f'(x)}{2 \sqrt{f(x)}} . \frac{2 \sqrt{x}}{1} \right]$ = $\displaystyle \frac{4}{2 \sqrt{9}} . \frac{2 \sqrt{9}}{1}$ = $\displaystyle 4 ........................Is \ this \ correct \ ? \ ? \ ? \ ? \ ? \ ?$
• Dec 8th 2009, 03:42 PM
mr fantastic
Quote:

Originally Posted by zorro
Here is what i have done

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

Using L'Hospitals rule

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

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

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

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

Yes.