Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
What happened to the 2 in the numerator? And if you sub in x=1 for the denominator it equals 0 which is undefined? By the way how did you post the maths equation with the right symbols? I tried copy and pasting the symbols from MS Word, but they didn't come out
no ... you can also think about it.
as $\displaystyle x \to 1$ , the numerator approaches $\displaystyle 1$
as $\displaystyle x \to 1$, the denominator approaches a very small positive value (why will it be positive?)
as a result ... $\displaystyle \frac{a \, fixed \, value}{a \, very \, small \, positive \, value} \to \infty$
Also, you can write
$\displaystyle \frac{2-x}{(x-1)(x-1)}=\frac{1+(1-x)}{(1-x)(1-x)}$
since (x-1)(x-1)=[-(1-x)][-(1-x)]=(1-x)(1-x)
leading to
$\displaystyle \frac{1}{(1-x)(1-x)}+\frac{(1-x)}{(1-x)(1-x)}$
$\displaystyle =\frac{1}{(1-x)(1-x)}+\frac{1}{1-x}$
As x tends to 1, both of these tend to $\displaystyle +\infty$
That's very true,
but the first one will be far more positive, giving an overall positive result.
If x=1.1
$\displaystyle \left(\frac{1}{-0.1}\right)\left(\frac{1}{-0.1}\right)+\frac{1}{-0.1}=\frac{1}{0.01}-\frac{1}{0.1}=100-10$
I should have said, as x tends to 1, the "sum" of these tends to $\displaystyle +\infty$ with the first fraction dominant.