Recent content by Joel Jacon

What is the difference between $a(t)$, $a(v)$ and $a(x)$? If $a(t) = \d{v}{dt}$ then what will $a(v)$ and $a(x)$ equal to? $a(t)$ is acceleration with change in time $a(v)$ is acceleration with change in velocity $a(x)$ is acceleration with change in position

I told that's what my teacher said. If I had known the meaning I wouldn't have asked the question.

I told that I didn't understand what the teacher told. Any link on web that explain it would be helpful. This is not a homework question.

What happen when the capillary rise occur in a tube of insufficient length? My teacher told me that hR = constant where h is height and R is radius of sphere of which the curved surface of meniscus firm a part. She also told me that if h become less so R has to increase so radius of meniscus...

What about the shape of meniscus?

Why is there excess pressure always on the concave side or surface of the meniscus? In my book it is also written that excess pressure balance the vertical resultant forces due to surface tension. How can a pressure balance a force? My teacher said that shape of meniscus does not depend on...

Yes, the question is correct. See the question 1 in the image

But the answer given in my book is -11. While using direct substitution I get 9. How can you get -11

Can anyone tell me how to solve the following limit by factorization method $\lim{{x}\to{5}} \frac{x^3 + 3x^2 - 6x + 2}{ x^3 + 3x^2 - 3x - 1}$?Please tell me how to factorize such big equation?

If modulus is the absolute value then can you explain me why we use $\lim_{{x}\to{0}} \frac{x}{|x|}$ as $\lim_{{x}\to{0}} \frac{x}{-x}$ for LHL and $\lim_{{x}\to{0}} \frac{x}{x}$ for RHL for the question in which we have to show $\lim_{{x}\to{0}} \frac{x}{|x|}$ does not exist

Can you please tell me how to deal with the modulus. i know that x - |x| is the function to be used for LHL as x<2

Evaluate $\displaystyle \lim_{{x}\to{2}} f(x$) if it exist where $f(x)$ = x - |x| where x<2;4 where x = 2;3x - 5 where x>2? LHL $\displaystyle \lim_{{x}\to{2}} f(2x)$ = 4 RHL $\displaystyle \lim_{{x}\to{2}} f(3x - 5)$ = 1 Therefore the limit x tend to 2 for the function does not exist...

I finally understand now. LHL is negative infinity and RHL is positive infinity.

I already know that as the denominator becomes smaller and smaller and closer to zero the value becomes infinity. Can you tell me how to get right hand limit and left hand limit for this problem.

Show that $\lim_{{x}\to{0}}$ $\frac{1}{x}$ does not exist? Please tell me how to make LHL and RHL for this? Explain me all steps used?