Help with solving limits Solve the following limits:
$$\lim _{x \to 3} \frac{1 - \sqrt{x-2}}{x^2 - 9} \quad \quad \text{and} \quad \quad \lim _{x \to \infty} \left(\frac{x^2-3}{x^2 + 7}\right)^{x^2 + x}$$

I'm relatively new to limits, but I managed to make some progress. Actually I solved the first limit, and conlcude that:
$$\lim _{x \to 3} \frac{1 - \sqrt{x-2}}{x^2 - 9} = -\frac{1}{12}$$ 
But I used L'Hopital Rule. Since L'Hopital Rule isn't allowed, is there any other way to solve it.
For the second one I managed to transform it to:
$$\lim _{x \to \infty} \left(1 - \frac{10}{x^2 + 7}\right)^{x^2 + x}$$
So i suspect the limit is:$\frac 1{e^{10}}$, because it looks very simular to the limit form for $e$ and I think it's due to the fact that $x^2 + x$ and $x^2 + 7$ grow simularly. But is there any better way to prove it, rather than just intuition.
P.S. I forgot to mention, but Taylor's Exapnsion is also not allowed.
 A: For the second one, note that:
$\log \left[\left(\dfrac{x^2-3}{x^2+7}\right)^{x^2+x}\right] = \dfrac{x^2+x}{x^2+7}\log\left (1 - \dfrac{10}{x^2+7}\right)^{x^2+7}$
Take the limit for each term in the product (using the continuity of $\log$ and the standard $e$ limit you refer to in your original post) and then exponentiate to obtain:
$\displaystyle\lim_{x\to \infty} \left(\dfrac{x^2-3}{x^2+7}\right)^{x^2+x} = e^{-10}$
A: For the first one, try multiplying by $\frac{1+\sqrt{x-2}}{1+\sqrt{x-2}}$ and then simplifying the resulting expression.
A: For the first limit, you can rationalize the numerator by multiplying the fraction by $\dfrac{1+\sqrt{x-2}}{1+\sqrt{x-2}}$. 
For the second limit, you have the correct answer. Other than l'Hopital's rule, I think you are left with your intuition, graphing, or evaluating lots of values of $x$ that grow larger to see what happens.
A: The previous answers give a good way to find the first limit, and I think you can find the second one by writing
$\big(\frac{x^2-3}{x^2+7}\big)^{x^2+x}=\bigg(\big(1-\frac{10}{x^2+7}\big)^{x^2+7}\bigg)^{\frac{1+\frac{1}{x}}{1+\frac{7}{x^2}}}\rightarrow \big(e^{-10}\big)^1=e^{-10}$
