Limits

[Guest]

hey, how would you do this type of question?

Show, using the properties of limites, that if lim (when x is approaching 5) f(x)=3, them lim (when x approaching 5) x^2-4/(f(x)=7

Thanks alot for the help

 

[Guest]

nvm I got it now you just put 3 into f(x) and then 5 for x and it equals to 7, but can anyone help me this question now?

FOR WHAT VALUE OF b DOES Lim (when x is approaching 1) x^2+bx-3/(x-1) exist?

 

[tkhunny]

Do you have a local teacher. If you're just substituting the value, exactly what you should NOT be doing, I think you may need a little personal review on the concept of a limit.

 

[galactus]

nvm I got it now you just put 3 into f(x) and then 5 for x and it equals to 7, but can anyone help me this question now?

FOR WHAT VALUE OF b DOES Lim (when x is approaching 1) x^2+bx-3/(x-1) exist?

Think about what value of b will factor the quadratic into \(\displaystyle (x-1)(x+?)\).
That way the (x-1)'s will cancel, otherwise, you'll be dividing by zero and that's a no-no.

 

[soroban]

Hello, bittersweet!

For what value of \(\displaystyle b\) does \(\displaystyle \lim_{x\to1} \frac{x^2\,+\,bx\,-\,3}{x\,-\,1}\) exist?

The limit will exist if \(\displaystyle x\,-\,1\) divides into \(\displaystyle x^2\,+\,bx\,-\,3\)

There is an algebraic way to work this out, but it's very messy!

If we're sharp, we can "eyeball" the problem and come up with the answer.

If \(\displaystyle b = 2\), we've got it!

We have: \(\displaystyle \lim_{x\to1}\frac{x^2\,+\,2x\,-\,3}{x\,-\,1} \;=\;\lim_{x\to1}\frac{(\sout{x\,-\,1})(x\,+\,3)}{\sout{x\,-\,1}}\:=\:\lim_{x\to1}(x\,+\,3)\:=\:4\)

 

[Guest]

huh tkhunny? so that isn't right what I did to solve that problem? and what do you mean if I have a local teacher? I have a teacher who teaches me at school..

Thanks for explaining the second question I posted on limits

 

[jolly]

huh tkhunny? so that isn't right what I did to solve that problem? and what do you mean if I have a local teacher? I have a teacher who teaches me at school..

Thanks for explaining the second question I posted on limits

 

[pka]

This is perfect theorem.
If each of f and g is a function such that
\(\displaystyle \L
\rm{lim}\limits_{x \to a} f(x) = K\) and
\(\displaystyle \L
\rm{lim}\limits_{x \to a} g(x) = L,\;L \not= 0\) then

\(\displaystyle \L
\rm{lim}\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{K}{L}\) .

That is what was used quite correctly.