I am new to calculus and I need to solve this question:
Find the limit of $$\frac{\sqrt{x^2+7}-\sqrt{x^3+3}}{\sqrt{x+1}-\sqrt{2x-1}}$$ as $x$ approaches $2$.
I tried rationalizing the equation but I always get an expression that returns $0/0$ when I substitute $x=2$ into the rationalized equation. Currently I got
$$(\sqrt{x^2+7}/\sqrt{x^3+x^2+7x+7}-(\sqrt{2x^3-x^2+14x-7})-\sqrt{x^3+3}/(\sqrt{x^4+x^3+3x+3}-\sqrt{2x^4-x^3+6x-3})) $$
but I am not sure how to proceed. I also tried viewing the numerator as f(x) and denominator as g(x) to apply the limit laws but that did not work out as well. Please help me and thanks in advance.
A good idea when dealing with $\frac 00$ limits with square root is to first rationalize the expression (note: I rearranged order, just so that we have positive quantities).
You got $f(x)=\Big(\sqrt{x^3+3}-\sqrt{x^2+2}\Big)\times\dfrac{\sqrt{x+1}+\sqrt{2x+1}}{x-2}$
Now it is always easier to figure out limits at zero rather than at a random point, so let substitute $x=2+u$ with $u\to 0$.
$f(u)=\underbrace{\Big(\sqrt{11+12u+6u^2+u^3}-\sqrt{11+4u+u^2}\Big)}_{O(u)}\times\underbrace{\dfrac{\sqrt{3+u}+\sqrt{3+2u}}{u}}_{\sim \frac{2\sqrt{3}}u}$
The first term has $\sqrt{11}-\sqrt{11}=0$, so it is basically something in $O(u)$ and it will compensate the $\frac 1u$ in the second term.
At this point, we can ignore terms of higher degree because they will be negligible, so the limit will be the same than the limit of
$$2\sqrt{3}\times\dfrac{\sqrt{11+12u}-\sqrt{11+4u}}{u}$$
As pointed out in the comment you can use the exapnsion $(1+u)^\alpha=1+\alpha u+o(u)$ or rewrite the expression to make appear $\frac{(1+u)^\alpha-1}u$ as suggested (if you are not comfortable with expansions).
In our case this will results in
$\require{cancel}\dfrac{2\sqrt{3}\sqrt{11}}{u}\Big((\cancel{1}+\frac 12.\frac{12}{11}u+o(u))-(\cancel{1}+\frac 12.\frac 4{11}u+o(u))\Big)=\dfrac{2\sqrt{3}\sqrt{11}}{u}\Big(\frac 4{11}u+o(u)\Big)\to\dfrac{8\sqrt{3}}{\sqrt{11}}$