1. Simple Limit problem

$\frac{49-x}{\sqrt{x}-7}$
$$\lim_{x \to 49}$$

$\frac{49-x}{\sqrt{x}-7}*\frac{\sqrt{x}+7}{\sqrt{x+7}}=\frac{49\sqrt{x}+ 343-x\sqrt{x}}{x-49}$

not sure where to go from here. I tried factoring out (x-49), but I still end up with an undefined answer.

What am I missing? (the answer is not undefined!)

Thanks!

2. $49 - x = (7)^2 - (\sqrt{x})^2 = (7 - \sqrt{x})(7+\sqrt{x})= -(\sqrt{x}-7)(7+\sqrt{x})$

or

use Lhopitals rule to get the derivatives of the numerator and the denominator.

3. Originally Posted by Vamz
$\frac{49-x}{\sqrt{x}-7}$
$$\lim_{x \to 49}$$

$\frac{49-x}{\sqrt{x}-7}*\frac{\sqrt{x}+7}{\sqrt{x+7}}=\frac{49\sqrt{x}+ 343-x\sqrt{x}}{x-49}$

not sure where to go from here. I tried factoring out (x-49), but I still end up with an undefined answer.

What am I missing? (the answer is not undefined!)

Thanks!

$(49-x)(\sqrt{x}+7)=49(\sqrt{x}+7)-x(\sqrt{x}+7)=49\sqrt{x}+49(7)-x\sqrt{x}-7x$

If x is 49, the numerator is 49(7)+49(7)-49(7)-7(49)

which means we've gotten nowhere!
Both numerator and denominator are still zero if x=49.

This is because we are still subtracting equal quantities in both numerator and denominator.

Notice how Harish's method leaves us without subtractions.

4. Originally Posted by Vamz
$\frac{49-x}{\sqrt{x}-7}$
$\lim_{x \to 49}$

$\frac{49-x}{\sqrt{x}-7}*\frac{\sqrt{x}+7}{\sqrt{x+7}}=\frac{49\sqrt{x}+ 343-x\sqrt{x}}{x-49}$

not sure where to go from here. I tried factoring out (x-49), but I still end up with an undefined answer.

What am I missing? (the answer is not undefined!)

Thanks!
Substitute $x = t^2$ and your limit becomes $\displaystyle \lim_{t \to 7} \frac{49-t^2}{t-7}$. The solution should be obvious.