Calculate  $\lim\limits_{ x\rightarrow 100 }{ \frac { 10-\sqrt { x } }{ x+5 } }$ $$\lim_{ x\rightarrow 100 }{ \frac { 10-\sqrt { x }  }{ x+5 }  } $$
Could you explain how to do this without using a calculator and using basic rules of finding limits?
Thanks
 A: We use the following limit laws:
$$\begin{align*}
&\lim_{x\to a}k = k\text{ if }k\text{ is constant}\tag{LoC}\\
&\lim_{x\to a}x = a\tag{LoV}\\
&\lim_{x\to a}(f(x)+g(x)) = \lim_{x\to a}f(x) + \lim_{x\to a}g(x)\text{ if both exist} \tag{LoS}\\
&\lim_{x\to a}(f(x)-g(x)) =\lim_{x\to a}f(x)-\lim_{x\to a}g(x) \text{ if both exist} \tag{LoD}\\
&\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\text{ if both exist and }\lim_{x\to a}g(x)\neq 0 \tag{LoQ}\\
&\lim_{x\to a}f(x) = f(a) \text{ if }f\text{ is continuous at }a \tag{Cont}
\end{align*}$$
("Limit of Constants", "Limit of the Variable", "Limit of Sums", "Limit of Differences", "Limit of Quotients", "Continuous function")
We also will use the fact that $f(x)=\sqrt{x}$ is continuous at every positive number.
Let's look at the denominator first:
$$\begin{align*}
\lim_{x\to100}(x+5) &= \lim_{x\to100}x + \lim_{x\to 100}5 &&\text{(by (LoS)}\\
&= 100 + 5 &&\text{(by (LoC) and (LoV)}\\
&= 105
\end{align*}$$
That means that the limit of the denominator exists, and is equal to $105$. Note in particular that it is not equal to $0$.
Now, the numerator: 
$$\begin{align*}
\lim_{x\to 100}(10-\sqrt{x}) &= \lim_{x\to 100}10 - \lim_{x\to100}\sqrt{x}&&\text{(by (LoD)}\\
 &= 10 - \sqrt{100} &&\text{(by (LoC) and (Cont)}\\
&= 10-10=0.
\end{align*}$$
Therefore, putting it all together, we have:
$$\begin{align*}
\lim_{x\to 100}\frac{10-\sqrt{x}}{x+5} &= \frac{\lim_{x\to 100}(10-\sqrt{x})}{\lim_{x\to 100}(x+5)} &&\text{by (LoQ)}\\
&= \frac{0}{105} &&\text{(by previous calculations)}\\
&= 0.
\end{align*}$$
A: Recall that $\lim_{x \to a} f(x) = f(a)$ if $f(x)$ is continuous at $a$. The function $f(x) = \dfrac{10-\sqrt{x}}{x+5}$ is continuous at $100$ since the numerator $10-\sqrt{x}$ is continuous for all $x > 0$ and the denominator $5+x$ is continuous for all $x$. Hence, the function $f(x) = \dfrac{10-\sqrt{x}}{x+5}$ is continuous for all $x > 0$. Now you should be able to finish it off.
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Hence, $$\lim_{x \to 100} \dfrac{10-\sqrt{x}}{x+5} = \dfrac{10 - \sqrt{100}}{100 + 5} = \dfrac{10 - 10}{105} = 0$$

A: I suppose that you asked this question not because it's a difficult question, but because you don't know very well the rules to take care of over the limits.
First of all you need to know what a limit
is, what the indefinite case are, and
why they are indefinite, what's the meaning behind this word (i.e. $ \frac{\infty}{\infty})$, and how to look things when facing a limit. You need to start learning basic things, and you may also play with them by using a computer to see the graph of a function when it takes certain values near the critical values 
you are looking for. I suppose the best way for you it's to receive an elementary explanation (this is possible) but i don't know what book i may recommend you for it. 
