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Limit Help lim as x approches -4 ((1/4)+(1/x)/(4+x)

Calculus1
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Does that make sense or should i draw it?
\[\huge \lim_{x \rightarrow -4}\frac{\frac{1}{4}+\frac{1}{x}}{4+x}\] It makes sense c: But I wanted to format it anway, heh.
yes that is it we have only covered limit laws so far.

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Other answers:

Ok this one isn't too bad.. it's just to test your silly math skills. So first let's get a common denominator on top.
My "silly" math skills are slowly coming back I took precalc 5 years ago
Remember how to get a common denominator? We'll just multiply them both together, that will be the easiest way to do it. So our common denominator will be 4x. It looks like the first term is missing an x, while the second term is missing a 4, So let's fix that. \[\huge \lim_{x \rightarrow -4}\frac{\left(\color{cornflowerblue}{\frac{x}{x}}\cdot \frac{1}{4}\right)+\left(\frac{1}{x}\cdot \color{cornflowerblue}{\frac{4}{4}}\right)}{4+x}\]Understand that part ok? :D
\[\huge \lim_{x \rightarrow -4}\frac{\left(\frac{4+x}{4x}\right)}{4+x}\]
okay so now can the denominator of the whole problem cancel out the numerator on the top leaving 4x?
oh well, um let's be careful a sec :)
We can write our problem like this,\[\large \lim_{x \rightarrow -4}\frac{\left(\frac{4+x}{4x}\right)}{4+x} \qquad =\qquad \lim_{x \rightarrow -4} \left(\frac{4+x}{4x}\right)\cdot \frac{1}{4+x}\]
Maybe this will help you see what's going on better. So what happens when we cancel those out? :O
it is 1/4x so the lim will be -1/16
Yay good job \c:/
thank you!

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