Trig limit: lim x -- > 1 [ sin ( x- 1 ) / x^2 + x - 2 ]

[grapz]

Hello, theres no answers for this in my text book, and i think this is a bit tricky problem so i'm looking for clarification.

lim x -- > 1 [ sin ( x- 1 ) / x^2 + x - 2 ]

lim x --> 1 [ sin ( x- 1 ) / ( x- 1) (x+2) ]

{ lim x-1 --> 0 [ sin ( x - 1 ) / ( x- 1 ) ] } * lim x --> 1 1 / x+ 2

1 x 1/3 = 1/3

 

[o_O]

Hmm never encountered a method such as changing \(\displaystyle \lim_{x \to 1} f(x)\) to \(\displaystyle \lim_{\left(x-1\right) \to 0} f(x)\). Anyway, if such a method is valid, then it looks correct. My graphing calculator seems to agree with your answer too: (.99999, .33333444) and (1.00001, .33333222).

 

[grapz]

Thats why i am not sure exactly.

But my reasoning is just taht as x --> 1, x - 1 will go to 0 so its technically a subsitutation

 

[o_O]

The last example from this site verifies this method is correct:

Let u = x - 1. Then as \(\displaystyle x \to 1\), then \(\displaystyle u \to 0\)

So:
\(\displaystyle \lim_{x \to 1} \left(\frac{sin(x-1)}{(x-1)(x+2)}\right)\)
\(\displaystyle = \lim_{u \to 0} \left(\frac{sinu}{u(u+3)}\right)\) (Since: x + 2 = (u + 1) + 2 = u + 3)
\(\displaystyle = 1 \cdot \frac{1}{3}\)
\(\displaystyle = \frac{1}{3}\)

And my graphing calculator verifies this so we can be pretty sure this is the correct limit!