# 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

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Just plug in: the function is continuous and defined at $100$. – Arturo Magidin Jun 28 '12 at 19:31
And you surely don’t need a calculator to find $\sqrt{100}$. – Brian M. Scott Jun 28 '12 at 19:31
There are many many solutions (very well written) below and no comments from the author! – Sigur Jun 28 '12 at 23:54

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*}

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thanks!! just what i needed! – Jiwon Jul 2 '12 at 0:18

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.

Move your mouse below the gray area for the answer.

Hence, $$\lim_{x \to 100} \dfrac{10-\sqrt{x}}{x+5} = \dfrac{10 - \sqrt{100}}{100 + 5} = \dfrac{10 - 10}{105} = 0$$

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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.

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