1. ## Limits involving infinity

Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1

2. Originally Posted by TsAmE
Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
hi
$\displaystyle \lim_{x\to 1}\left (\frac{x}{(x-1)^2} \right )=\infty$

3. Originally Posted by TsAmE
Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
It has no limit, as x approaches 1,
since the graph approaches infinity.

4. 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

5. Originally Posted by TsAmE
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
$\displaystyle \lim_{x \to 1} \frac{2-x}{(x-1)^2} = +\infty$.

6. Sigh I can't see the pictures, its says the following: To view links or images in signatures your post count must be 10 or greater. You currently have 3 posts.

7. look at the graph ...

8. Thanks a lot. So is the only way to solve this problem by looking at the graph, as you cant cancel out the denominator?

9. Originally Posted by TsAmE
Thanks a lot. So is the only way to solve this problem by looking at the graph, as you cant cancel out the denominator?
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 ... $\displaystyle \frac{a \, fixed \, value}{a \, very \, small \, positive \, value} \to \infty$

10. Originally Posted by TsAmE
Hello could someone please help me with this question: Find the limit of 2-x / (x - 1)^2 as x-> 1
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)

$\displaystyle \frac{1}{(1-x)(1-x)}+\frac{(1-x)}{(1-x)(1-x)}$

$\displaystyle =\frac{1}{(1-x)(1-x)}+\frac{1}{1-x}$

As x tends to 1, both of these tend to $\displaystyle +\infty$

11. But if you sub in x = 1.11111111 for the 1/ 1 - x part, the limit becomes negative

12. Originally Posted by TsAmE
But if you sub in x = 1.11111111 for the 1/ 1 - x part, the limit becomes negative
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.

13. Oh ok I see makes sense. So if I were to do this question in a test without a calculator, how would I know that I must test values close to the limit? Must I do this if I cant seem to cancel out the denominator?