#Help computing limits (intermediate form)
56 messages · Page 1 of 1 (latest)
- Wait patiently for a helper to come along.
- Once someone helps you, say thank you and close the thread with:
+close
- Feel free to nominate the person for helper of the week in #helper-nominations
- Do not ping the mods, unless someone is breaking the rules.
- If you're happy with the help you got here, and the server overall, you can contribute financially as well:
What?
I was suggesting using the literal definition of derivation
The derivative is the limit at a given point
But you cannot use l'hospital rules
yeah I mentioned that and I already used l' "hospital" rule and got the answer but I can not use it to solve
thanks for helping though
nope
Let me get into it
I guess multiplying by the conjugates of the numerator once and the denominator
haha yeah I was having it on my mind earlier but I said why not to seek for some help
lemme try it and see
Send it i am interested
the thing is i got other limits probs that i cant solve with l'hopital's
fine
Send
And @ me
@cosmic lava
ay man srry for late but im still a latex beginner
,tex $\lim_{x\to 1}\frac{\sqrt{2x-1}-1}{\sqrt{3x+1}-2} \[0.5cm] \frac{\sqrt{2x-1}-1}{\sqrt{3x+1}-2} \cdot \frac{\sqrt{2x-1}+1}{\sqrt{2x-1}+1} \cdot \frac{\sqrt{3x+1}+2}{\sqrt{3x+1}+2}$ \ Evaluating all this results in $\frac{(2x-2)\cdot \sqrt{3x+1}+2}{(3x-3)\cdot \sqrt{2x-1}+1} = \frac{2(x-1)\cdot \sqrt{3x+1}+2}{3(x-1)\cdot \sqrt{2x-1}+1} = \frac{2 \cdot \sqrt{3x+1}+2}{3 \cdot \sqrt{2x-1}+1 }$ \ Then plotting one will give us $\frac{4}{3}$
Halversen
limits like this
Factorize
How do you come to 4/3
huh?
plotting one will give you 8/6 which will equal 4/3
if you mean binomial theorem, i cant use that either
Can you use maths
,w binome de newton
Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online
,w newton binom
Fuck wolfram
my teacher's sweating me
yeah bro i cant use that, that's the binomial theorem
i only get to use one formula
$lim_{x\to a} \frac{x^{m}-a^{n}}{x^{m}-a^{n}}=\frac{m}{n} \cdot a^{m-n}$
Halversen
i dont even know what this formula called and i tried to search online i found nothing
so basically im trying to make the nominator to be x-2 with whatever exponents on x or 2
+close