Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$\lim_{ x\rightarrow 100 }{ \frac { 10-\sqrt { x } }{ x+5 } } $$

Could you explain how to do this without using a calculator and using basic rules of finding limits? Thanks

share|cite|improve this question
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: $$\begin{align*} \lim_{x\to 100}\frac{10-\sqrt{x}}{x+5} &= \frac{\lim_{x\to 100}(10-\sqrt{x})}{\lim_{x\to 100}(x+5)} &&\text{by (LoQ)}\\ &= \frac{0}{105} &&\text{(by previous calculations)}\\ &= 0. \end{align*}$$

share|cite|improve this answer
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, $$\lim_{x \to 100} \dfrac{10-\sqrt{x}}{x+5} = \dfrac{10 - \sqrt{100}}{100 + 5} = \dfrac{10 - 10}{105} = 0$$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.