limit, as x approaches 3, of (sqrt(3x) - 3])/ (2x - 6)

[xc630]

Hello, I'm absolutely terrible at finding limits without a calculator. Does anyone have any general tips? For instance I was trying to find the limits of these two problems.

lim of (SQRT(3x)-3) / (2x-6) as x approaches 3.

Here I know you can't plug in 3 and when I tried manipulating the limit algebraically by multiplying by the conjugate I got no where.

Also for the lim of x ^ (98/(1-x)) i do not know where to start. Thanks for any help

 

[galactus]

Re: limits

On the first one, yes you can use the conjugate of the numerator.

\(\displaystyle \lim_{x\to{3}}\frac{\sqrt{3x}-3}{2x-6}\)

Now, multiply the topo and bottom by \(\displaystyle \sqrt{3x}+3\)

Upon doing that we get:

\(\displaystyle \lim_{x\to{3}}\frac{3(x-3)}{2(x-3)(\sqrt{3x}+3)}\)

Take note that the (x-3)'s cancel and we have:

\(\displaystyle \lim_{x\to{3}}\frac{3}{2(\sqrt{3x}+3)}\)

Now, let x approach 3 and we get:

\(\displaystyle \frac{3}{2(\sqrt{9}+3)}=\frac{1}{4}\)

See how to do that now?.

As for the last one, what does x approach?. Is that \(\displaystyle x^{\frac{98}{1-x}}\)

 

[xc630]

Re: limits

Thank you for the first one. I had multiplied the bottom to be 2x SQRT(3x) +6x - 6SQRT(3x) -18.

For the second one x approaches 1

 

[pka]

Re: limits

For the second one x approaches 1

\(\displaystyle \begin{array}{l} y = x^{\frac{9}{{1 - x}}} \quad \Rightarrow \quad \ln (y) = \frac{{9\ln (x)}}{{1 - x}}\mathop = \limits^H \frac{9}{{ - x}} \to - 9 \\ y \to e^{ - 9} \\ \end{array}\)

 

[galactus]

Re: limits

Do you know L'Hopital's rule?. I usually refrain from using it, but........

\(\displaystyle \lim_{x\to{1}}x^{\frac{98}{1-x}}\)

Rewrite:

\(\displaystyle \lim_{x\to{1}}e^{98\frac{ln(x)}{1-x}}\)

\(\displaystyle e^{98\lim_{x\to{1}}\frac{ln(x)}{1-x}}\)

Now, using L'Hopital, we get:

\(\displaystyle e^{98\lim_{x\to{1}}\frac{-1}{x}}\)

As x approaches 1, we are left with \(\displaystyle e^{-98}=\frac{1}{e^{98}}\)

If you don't know L'Hopital, perhaps we can work around that. It was just an easy out in this event.

 

[xc630]

Re: limits

I know the form f (x)/ g(x) = lim x approaches c f'(x)/ g' (x)

I'm confused about where the e power comes into play

 

[galactus]

Re: limits

e is used because seeing the x with an exponent we recognize that

\(\displaystyle e^{\frac{98ln(x)}{1-x}}=x^{\frac{98}{1-x}}\)