You cant

You cant.

Other urls found in this thread:

wolframalpha.com/input/?i=lim (x-3)/ln(x - 2) as x->3-
en.wikipedia.org/wiki/Newton's_method
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You're right

That's a pretty easy one there mate maybe it's time you start doing your homework
>t. Math uni graduate

The thing is my solution is 1, but the teacher's solution is 0. That's why I posted it here.

You can't at 3, but you're taking the limit as it approaches 3 from the left.

Quite easy tho

So it's 0?

Its zero because x approaches 3, numerator goes to zero. You could do la'hopital (idr how to spell it) if you need confirmation

I did it, idk why it's 0

Nothing in that equation is undefined at 3.

ln(1) is also 0. Wolfram Alpha says the answer is 1.

Yes, both numerator and denominator goes to 0, that's why I did de L'Hospital you can't divide 0 by 0

Your teacher is wrong and you are right, L' Hospital's to get 1/(1/x-2) plug in 3 to get one.


wolframalpha.com/input/?i=lim (x-3)/ln(x - 2) as x->3-

T. Engineering Student

Thanks. :)

hospital. easy. 1.

As I near the end of my first sem of calc, I am proud to say I actually can!

f(x) = 0.5x + sin(x)
>find number and postion of zeroing

when you're done with maths you'll see the solution just by looking at it. of course it's 1 because close to x=3 both functions have increasingly similar values without one function "beating". your prof basically said, that around x=3 the upper function is getting close to 0 faster and the lower one can't keep the pace. if the lower function would reach 0 faster than the upper one the limit would be oo in many cases.

>f(x) = 0.5x + sin(x)
Just one zero point, and it is at x=0.

k i fucked that up i guess. term seems to be only for x axis then. non native...
f(x) = 0..

and again. f(x) = 0.5x - sin(x)
fuug

What about this one? My solution is 1 divided by 0 (1/0) so there is no answer, when again my teacher's solution is 0?

Y'all forgot the limit
X tends to 1
Denom goes to 0
Answer is 0

It's a 0/0 case, so you must use L'Hopital's rule and then take the limit

it's 0

...= -inf + lim(18x^2 / sqr(9x^2 + 1)

...

...

OK that's a bit better, there are three (or at most five, I'm not sure yet) roots just by looking at the equation. One is the "old" x=0, the other ones are symmetric around the y axis, somewhere around +-pi/2 (and maybe around +-3pi/2). Sin value goes just from -1 (at 3*pi/2 + 2*k*pi) through zero (at whole pi multiples) to 1 (at pi/2 + 2*k*pi) for each whole k. 0.5x would 'overpower' sin from x>2 and x

lim = 1

done. good job user.

en.wikipedia.org/wiki/Newton's_method
for example.

Kys, you gotta spell lim on the beginning of each term, the "lim" goes away when you replace the values

Try something a little more difficult.

>how to wolfram alpha