- #1

- 24

- 0

can someone plzzzzz show me the simplist way to do it

thx

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mooneh
- Start date

- #1

- 24

- 0

can someone plzzzzz show me the simplist way to do it

thx

- #2

- 158

- 0

you want to find limit when x tends to a

substitute x with a+h

now find the directive limit for h tends to 0

you see that only difference in both limits is the sign of h,value remain same

so take underconsideration the sign and substitute 0 in function if it is defined for both sides

- #3

- 24

- 0

can u give me an example ?

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 964

Here's one easy example:

[tex]\lim_{x\rightarrow 1^+} x^2[/tex]

Since [itex]x^2[/itex]

[tex]\lim_{x\rightarrow 1^+} x^2= \lim_{x\rightarrow 1^-} x^2= 1[/tex]

Here's a slightly harder example:

[tex]\lim_{x\rightarrow 1^+} f(x)[/tex]

where f(x)= [itex]x^2[/itex] if x< 1 and if f(x)= x+ 4 if x> 1.

Of course, [itex]\lim_{x\rightarrow 1^+} f(x)[/itex] depends

[tex]\lim_{x\rightarrow 1} x+ 4[/itex]

which is 5.

[tex]\lim_{x\rightarrow 1^+} f(x)= 5[/tex]

Similarly

[tex]\lim_{x\rightarrow 1^-} f(x)= \lim_{x\rightarrow 1} x^2= 1[/tex]

In this case, since the two "one-sided" limits are different, the "limit" itself does not exist. Typically, you find one-sided limits

- #5

- 158

- 0

[tex]lim[/tex][tex]\sqrt{1-x}[/tex]

[tex]x\rightarrow 1[/tex]

find right hand limit ,it is undefined because , if you by making x=1+h

then you see that root of negative no does not exist but left hand limit does exist and is 0

[tex]x\rightarrow 1[/tex]

find right hand limit ,it is undefined because , if you by making x=1+h

then you see that root of negative no does not exist but left hand limit does exist and is 0

Last edited:

Share: