Limits involving infinity

**TsAmE** · April 18th 2010, 04:27 AM

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

**Raoh** · April 18th 2010, 04:44 AM

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

$\lim_{x\to 1}\left (\frac{x}{(x-1)^2} \right )=\infty$

**Archie Meade** · April 18th 2010, 04:48 AM

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.

**TsAmE** · April 18th 2010, 04:51 AM

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

**mr fantastic** · April 18th 2010, 04:58 AM

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

$\lim_{x \to 1} \frac{2-x}{(x-1)^2} = +\infty$ .

**TsAmE** · April 18th 2010, 06:54 AM

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.

**skeeter** · April 18th 2010, 07:48 AM

look at the graph ...

**TsAmE** · April 18th 2010, 08:08 AM

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

**skeeter** · April 18th 2010, 08:41 AM

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 $x \to 1$ , the numerator approaches $1$

as $x \to 1$ , the denominator approaches a very small positive value (why will it be positive?)

as a result ... $\frac{a \, fixed \, value}{a \, very \, small \, positive \, value} \to \infty$

**Archie Meade** · April 18th 2010, 09:18 AM

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

$\frac{2-x}{(x-1)(x-1)}=\frac{1+(1-x)}{(1-x)(1-x)}$

since (x-1)(x-1)=[-(1-x)][-(1-x)]=(1-x)(1-x)

leading to

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

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

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

**TsAmE** · April 18th 2010, 01:19 PM

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

**Archie Meade** · April 18th 2010, 01:41 PM

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

$\left(\frac{1}{-0.1}\right)\left(\frac{1}{-0.1}\right)+\frac{1}{-0.1}=\frac{1}{0.01}-\frac{1}{0.1}=100-10$

I should have said, as x tends to 1, the "sum" of these tends to $+\infty$ with the first fraction dominant.

**TsAmE** · April 19th 2010, 04:40 AM

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?

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