# Indeterminate Limits

I have been studying independently through various online courses and I still have trouble understanding what to do with certain limits. I am hoping for some guidance on the following two problems to help me solve them (I do not need to answer as much as help understanding where I am going).

$$\lim_{x\to 0} \frac{\sqrt{16 + 4x} - \sqrt{16 - 4x}}{x}$$

$$\lim_{x\to 0} \frac{\frac{1}{(x + 6)^2} - \frac{1}{36}}{x}$$

I can tell that these are both $\frac{0}{0}$ indeterminate equations and I can simplify them in a number of ways but I cannot seem to get the 0 out of the bottom of the equation. Any help pointing me in the right direction would be greatly appreciated!

• I suggest you use \lim_{x \to a} for limits. You can go here for further help with typesetting common math expressions. – Zev Chonoles Jun 1 '13 at 22:48
• Thank you! I wasn't sure how that was handled :) – Tim C Jun 1 '13 at 22:49

For the second one, you may find the difference of squares formula $(a+b)(a-b)=a^2-b^2$ useful, together with the observations that $$\frac1{(x+6)^2}=\left(\frac1{x+6}\right)^2$$ and $$\frac1{36}=\left(\frac16\right)^2.$$ Use that to rewrite the numerator and see what happens!

Surprisingly enough, we can once again use the difference of squares for the (fixed version) of your first one! We'll be going the other way, though. For $x$ sufficiently close to $0$, we have that $\sqrt{16-4x}$ and $\sqrt{16+4x}$ are both positive, so that in particular, $$\sqrt{16+4x}+\sqrt{16-4x}\ne 0.$$ From there, the difference of squares formula tells us that $a-b=\frac{a^2-b^2}{a+b}$ so long as $a+b\ne 0$. This lets us in particular rewrite $$\sqrt{16+4x}-\sqrt{16-4x}=\frac{(16+4x)-(16-4x)}{\sqrt{16+4x}+\sqrt{16-4x}}=\frac{8x}{\sqrt{16+4x}+\sqrt{16-4x}}.$$ Can you take it from there?

P.S.: You may also find my answer here to be helpful, dealing as it does with the more general case.

I would assume that you typed in the expression for the limit correctly.

For the problem you provided, there is a short way to determine the limit. Given that the limit is:

$$\lim_{x\to 0} \frac{\sqrt{16 + 4x} - \sqrt{16 + 4x}}{x}$$

The numerator expression becomes $0$. Then, we have:

$$\lim_{x\to 0} \frac{0}{x} = \lim_{x \to 0} 0 = 0$$

~~

For the second problem, we can rewrite the expression as:

$$\lim_{x \to 0} \dfrac{\frac{1}{(x + 6)^2} - \frac{1}{36}}{x} \cdot \dfrac{36(x + 6)^2}{36(x + 6)^2}$$ $$= \lim_{x \to 0} \dfrac{36 - (x + 6)^2}{36x(x + 6)^2}$$ $$= \lim_{x \to 0} \dfrac{36 - x^2 - 12x - 36}{36x(x + 6)^2}$$ $$= \lim_{x \to 0} \dfrac{-x(x + 12)}{36x(x + 6)^2}$$ $$= \lim_{x \to 0} \dfrac{-(x + 12)}{36(x + 6)^2}$$

Thus,

$$\lim_{x \to 0} \dfrac{-(x + 12)}{36(x + 6)^2} = \dfrac{-12}{36(6)^2} = -\dfrac{1}{108}$$

Verified by Wolfram for:

First problem

Second problem

For the problem you edited

The given limit is:

$$\lim_{x \to 0} \dfrac{\sqrt{16 + 4x} - \sqrt{16 - 4x}}{x}$$

Multiply the top and bottom by the conjugate of the numerator expression, which is $\sqrt{16 + 4x} + \sqrt{16 - 4x}$. This gives:

$$\lim_{x \to 0} \dfrac{16 + 4x - (16 - 4x)}{x(\sqrt{16 + 4x} + \sqrt{16 - 4x})}$$ $$= \lim_{x \to 0} \dfrac{8x}{x(\sqrt{16 + 4x} + \sqrt{16 - 4x})}$$ $$= \lim_{x \to 0} \dfrac{8}{\sqrt{16 + 4x} + \sqrt{16 - 4x}}$$

Thus,

$$\lim_{x \to 0} \dfrac{8}{\sqrt{16 + 4x} + \sqrt{16 - 4x}}$$ $$= \dfrac{8}{\sqrt{16} + \sqrt{16}}$$ $$= \dfrac{8}{4 + 4}$$ $$= 1$$

Verified by WolframAlpha.

Since you said that you have trouble understanding certain indeterminate limits, let me just give you some general tips on how to solve them. These are the basics tools you should always apply when appropriate. When you have a sum or difference in your limit (especially involving square roots and trig functions, since you can often get $sin^2x+cos^2x$)) it is usually VERY useful to multiply by the conjugate (even if your denominator is $1$). With limits involving trigonometric functions, always look for some special limits, most commonly the limit as $x$ approaches $0$ of $sin(x)/x=1$. If you have a mix of algebra and trig, you can try substitutions such as (we got this from the above limit) $x=sin(x)$ (iff $x$ is approaching $0$). With exponential, try to look for the definition of $e$, which is the limit as $x$ approaches infinity of $(1+1/x)^x$, or equivalently the limit as $y$ approaches $0$ of (1+y)^(1/y) (note that here we just made the substitution $y=1/x$; this is also a very common and useful tool). Lastly, come the limits which are disguised as differentiation problems, such as lim as $x$ approaches $0$ of $(cos(x)-1/x)$. This can be rewritten as $(cos(x)-cos(0))/x$, so this is just the derivative of cosine at x=0. Don't forget how useful substitution is in unmasking certain limits! For example, the limit as $x$ approaches infinity of $x*sin(1/x)$. This might seem hard, but if we let $t=1/x$ this becomes the limit as t approaches 0 of $sin(t)/t$, which is a special limit equal to $1$.