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simple limit but i am blank

Mathematics
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\[\huge \lim_{x \rightarrow 5} \frac{ 1 }{ x-5 }\]
Does it exist?
yes

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Other answers:

Are you sure about that?
not really
i completely forgot how to solve these except for direct substitution
Well, when you can't directly substitute, I recommend putting in really close values (e.g. 5.001 and 4.999).
Make sure it exists first. Then you can use things like squeeze theorem / l'Hospital's rule. There is no sure way of doing limits, only a series of methods.
okay thank you :)
how would i solve limits that have fractions over fraction though ?
Start by simplifying it into a single fraction.
\[\large \lim x \rightarrow -3 \frac{ \frac{ 1 }{ x } +\frac{ 1 }{ 3 }}{x+3}\]
ohh okay
(a/b)/c = a/(bc) and a/(b/c) = (ac)/b
Okay you need to add up the \(1/x\) and \(1/3\).
a/b + c/d = (ad + bc)/(bd)
\[\large \lim \rightarrow -3 \frac{ 3+x }{ 3x^2+9x }\]
I don't think you're doing it correctly. \[ \frac{1}{x} +\frac{1}{3} = \frac{x+3}{3x} \]And then \[\Large \frac{\frac{x+3}{3x}}{x+3} = \frac{x+3}{3x(x+3)} = \frac{1}{3x} \]
This function becomes continuous once you have manipulated it a bit.
Continuous at \(-3\) at least.
what does that mean ?
What does continuous mean? It means: 1) \(f(a)\) is defined 2) \(\lim_{x \to a}f(x)\) exists and 3) \(\lim_{x \to a}f(x) = f(a) \) In general, it means you can just plug \(a\) into \(f(x)\) and get the answer to the limit.
ohh thankyou !
Ask yourself, what is a limit? Really. It's what happens as you approach a number from both sides. So to approach a number, you can plug in numbers close to it, like 4.999 is close to 5 on the left side while 5.00001 is close on the right side. Make sense? Plugging these in and seeing what you approach is really the essence of a limit and understanding that will let you solve any limit problem by simply graphing it when you get stumped.

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