Question about taking limits in calculus

I am working on the final exam review for a college level calculus class. Right now, I am dealing with the Limits Section. Some of the limits are easy to find, but some are giving me trouble.

Here are some of the problems

1) Find the limit as X approaches 4 (Square root of X -2)/(x-4). The Square root in the numerator only contains the X, not the 2. If I plug in the 4 into the equation, I will get a zero on the top and bottom, which is Undefined. However, the answer in the guide is 1/4 How do they get it?

2) Find the limit as X approaches 4 of (1/x)-(1/4)/(x-4). Again, same problem. I get zero if I plug 4 into it? What am I doing wrong here. It works for some limit problems. By the way, they got -1/16

3) Find the limit as X approaches infinity of (2x-3)/The Square Root of (4x^2 +5). Completely lost here. Answer they gave was 1

4) Lastly, as X approaches Infinity of (x-Square root of (x^2+4)). The whole x^2 + 4 is under the square root symbol. They got 0 as the answer.

I appreciate any help you are able to provide. I'm trying really hard to ace this test and I need to understand everything in this packet.

alright, so lets take a look at these. your main issue here is not realizing (or not knowing) which rule to apply here. in this case we are going to use L'Hoptials Rule. This states that when asked for the limit as x approaches some value of function f(x)/g(x), that limit can also be evaluated by taking that same limit of f'(x)/g'(x). f'(x) is the derivative of f(x) and same for g(x).

1) so lets take the top function, sqrt(x)-2. the -2 portion is irrelevant because the derivative of a constant is 0, so we are just going to differentiate the sqrt(x). the derivative of sqrt(x) is 1/(2sqrtx).
the derivative of the bottom is just 1 seeing as the coefficient of x is just . so we're just left with 1/(2sqrt(x)) again. plugging in 4 you get 1/(2*2) which is 1/4