simple limit but i am blank

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

simple limit but i am blank

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question