One sided limits?

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Jul 14th 2011, 01:53 PM
homeylova223

One sided limits?

Find the limit if it exists

1. As delta x approaches 0 limit from the left

((1/x+delta(x)-(1/x))/((delta(x))

2. This second limit is kind of confusing

As x approaches 3

where f(x)=

(x+2)/2 x < or equal to 3

12-2x/3 x>3

Would I just plug in 3+2/2=2.5?
Jul 14th 2011, 02:13 PM
Plato

Re: One sided limits?

Quote:

Originally Posted by homeylova223

Find the limit if it exists
1. As delta x approaches 0 limit from the left
((1/x+delta(x)-(1/x))/((delta(x))

Is this the problem $\displaystyle\lim _{\delta x \to 0^ - } \dfrac{{\frac{1}{{x + \delta x}} - \frac{1}{x}}}{{\delta x}}~?$

Please learn to post readable questions.
Jul 14th 2011, 02:15 PM
homeylova223

Re: One sided limits?

Yes it is like that. I should learn that latex.
Jul 14th 2011, 02:23 PM
Siron

Re: One sided limits?

Determine for the fractions in the numerator a common denominator.
Jul 14th 2011, 02:23 PM
Plato

Re: One sided limits?

Quote:

Originally Posted by homeylova223

Yes it is like that. I should learn that latex. PLEASE DO

$\frac{{\frac{1}{{x + \delta x}} - \frac{1}{x}}}{{\delta x}} = - \frac{{\delta x}}{{\delta x\left[ {x\left( {x + \delta x} \right)} \right]}}$
Jul 14th 2011, 09:11 PM
mr fantastic

Re: One sided limits?

Quote:

Originally Posted by homeylova223

Find the limit if it exists

1. As delta x approaches 0 limit from the left

((1/x+delta(x)-(1/x))/((delta(x))

[snip]

This has the form of a derivative from first principles, where f(x) = ...... and so the answer is f'(x) = ......

Quote:

Originally Posted by homeylova223

Find the limit if it exists

[snip]

2. This second limit is kind of confusing

As x approaches 3

where f(x)=

(x+2)/2 x < or equal to 3

12-2x/3 x>3

Would I just plug in 3+2/2=2.5?

The right hand limit is 12 - 2(3)/3 = .... and the left hand limit is (3 + 2)/2 = ....

If right and left hand limits are the same then the limit exists and is equal to .... If right and left hand limits are not the same then the limit does not exist.