Determining $\lim_{x \to -25} (\sqrt{x} + 5)/(x - 25) = \lim_{x \to -25} 1/(\sqrt{x} - 5)$

Question

I have a maths problem but the solution I have been given does not look correct. The problem is as follows:

$$\lim_{x \to -25} \frac{\sqrt x +5}{x-25}$$

The solution proceeds to factorise the denominator.

$$\lim_{x \to -25} \frac{\sqrt x +5}{\left( \sqrt x -5 \right)\left( \sqrt x +5 \right)}$$

At this point the solution ends and says that the limit is $\infty$.

However, I can see how to get to this. If I continue to simplify the limit equation I get:

$$\lim_{x \to -25} \frac{1}{\sqrt x -5}$$

But if I now substitute, I will end up with square root of a negative number so I don't see how end up with infinity. Can anyone help?

Answer 1

There seems to be something wrong with the problem. The first line already contains a square root of a negative number. Factorising will not help with that.

The solution you give does work for $\lim_{x\to +25}$ instead of $\lim_{x\to -25}$.

Answer 2

As already noticed the limit you are looking for is meaningless since the expression is not defined for $x<0$. Maybe you are looking for

$$\lim_{x \to 25} \frac{\sqrt x -5}{x-25}$$

which is in the form $\frac 0 0$.

In this case you mehod is fine and we obtain

$$\lim_{x \to 25} \frac{\sqrt x -5}{\left( \sqrt x -5 \right)\left( \sqrt x +5 \right)}=\lim_{x \to 25} \frac{ 1}{ \sqrt x +5 }=\frac 1 {10}$$

Note that for the limit

$$\lim_{x \to 25} \frac{\sqrt x +5}{x-25}$$

we can conclude that it doesn't exist since the numerator tends to $10$ but the denominator tends to $0$ therefore the expression diverges at $\pm \infty$ depending on the sign of $x-25$.