Find the limit of the function:
(tanp)(x-1)/(tanq)(x-1)
as x approaches 1. We're hinted at using the result that sinx/x = 1 as x approaches 0.
Im not sure how we go abt doing this qs.
thanks in advance
as written the answer is simply tanp/tanq
what is tanp?
what is tanq ?
what is p and what is q ?
Does this mean tan (px - p) / tan (qx - q)? or something?
tanp and tanq is short for tangent p and tangent q respectively(we havent been told where p is a variable or constant; same goes for q)
no it doesnt mean tan (px - p)
it wud be xtanp -tanp if u were to multiply it through ..same for denominator
and the answer given is p/q
(xtanp-tanP)/(xtanq-tanq) as x approachtes zero.
Unless p,q are somehow related to x, it is not equal to p/q.
to get p/q, it looks like somebody canceled the "tan"
reminded me of the time when one of my students gave this solution to
Lim (sinx)/x as x = 0
= sin
I told him to sin no more, but he didn't get it.
To find the limit of the given function as x approaches 1, we can start by simplifying the expression:
Let's rewrite the function using the given hint that sinx/x = 1 as x approaches 0:
(tanp)(x-1)/(tanq)(x-1) --> (tanp)(sin(x-1)/(x-1))/(tanq)(sin(x-1)/(x-1))
Now, let's focus on the expression sin(x-1)/(x-1). As x approaches 1, x-1 approaches 0. So, we can use the given result sinx/x = 1 as x approaches 0 to evaluate this expression:
sin(x-1)/(x-1) --> 1 as x approaches 1
Plugging this result back into the original function, we have:
(tanp)(sin(x-1)/(x-1))/(tanq)(sin(x-1)/(x-1)) --> (tanp)(1)/(tanq)(1)
Since tanp and tanq are constant values (not dependent on x) and any number divided by 1 is itself, the limit of the function as x approaches 1 is:
(tanp)/(tanq)
Therefore, the limit of the given function as x approaches 1 is (tanp)/(tanq).