Help with simplifying a limit function.

coooool222

Junior Member
Joined
Jun 1, 2020
Messages
93
Hello I need help with this problem, they say that its -1. I looked it up and said I have to use derivatives but Im still on limits. We have not started derivatives yet so I have to simplify this algebraically. I am getting 0

1663344042800.png

(x-b)49-x+b =
0
 
Hello I need help with this problem, they say that its -1. I looked it up and said I have to use derivatives but Im still on limits. We have not started derivatives yet so I have to simplify this algebraically. I am getting 0

View attachment 34052

(x-b)49-x+b =
0
What you're actually getting is not 0, but 0/0, which is indeterminate. That means you have to use a different method than just substituting.

See if you can factor (x-b) out of the numerator, in order to cancel it. Cancellation is a typical goal in limit problems.
 
It would help if you showed a bit more of your working. I think you made a simple mistake in algebra.

It is I find frequently helpful to forget about the behaviour of the function at the limiting value initially. After all, the limit specifically does not include that value BY DEFINITION.

xb and y=(xb)50x+bxb=(xb)50(xb)(xb)=(xb){(xb)491}(xb)(xb)49x+b. x \ne b \text { and } y = \dfrac{(x - b)^{50} - x + b}{x - b} = \dfrac{(x - b)^{50} - (x - b)}{(x - b)} = \\ \dfrac{(x - b)\{(x - b)^{49} - 1\}}{(x - b)} \ne (x - b)^{49} - x + b.
 
(x-b)49-x+b =
0
I missed, before, the fact that you were not just evaluating the numerator, but making a faulty attempt to simplify.

You apparently just cancelled the (x-b) in the first term, but left the second term untouched, forgetting that both terms in the numerator are being divided by (x-b).

But my suggestion still stands: you need to factor the numerator, so that you can cancel a factor from the entire thing, not just one term. Alternatively, you can just divide each term by (x-b).
 
Consider 5+255=305=6\displaystyle \dfrac{5+25}{5}=\dfrac{30}{5}=6. Now let's try it your way. 5+255\displaystyle \dfrac{5+25}{5}=1+25 =26 (I reduced 5/5 to 2). Which is correct?

As Dr Peterson pointed out, you need to divide both factors in the numerator by 5, ie 5+255=55+255\displaystyle \dfrac{5+25}{5}=\dfrac{5}{5} +\dfrac{25}{5}=1 + 5 = 6
 
Top