1. ## Confusion with limits

$\displaystyle \lim_{x \to -3} \frac{t^2-9}{2t^2 +7t+3}$ will equal 0 if you apply the usual limit laws and consider the limit as t tends to -3 in the numerator.

Factorisation of the numerator and denominator gives us $\displaystyle \frac{t-3}{2t+1}$, which gives us the correct limit $\displaystyle \frac{6}{5}$. However, I only know this because I looked at the graph.

It seems arbitrary that a simplification should produce a different result, if the expression is equal. How can I possibly tell whether a function has a limit at a particular point unless I do the "correct" manipulations? There must be some strict method I can apply to every function to ensure I know I am taking limits correctly.

2. Originally Posted by rowe
$\displaystyle \lim_{x \to -3} \frac{t^2-9}{2t^2 +7t+3}$ will equal 0 if you apply the usual limit laws and consider the limit as t tends to -3 in the numerator.

Factorisation of the numerator and denominator gives us $\displaystyle \frac{t-3}{2t+1}$, which gives us the correct limit $\displaystyle \frac{6}{5}$. However, I only know this because I looked at the graph.

It seems arbitrary that a simplification should produce a different result, if the expression is equal. How can I possibly tell whether a function has a limit at a particular point unless I do the "correct" manipulations? There must be some strict method I can apply to every function to ensure I know I am taking limits correctly.
The function $\displaystyle \frac{t^2-9}{2t^2 +7t+3}$ is not defined at $\displaystyle x=-3$ because the denominator is zero. You can't have a zero denominator. When you set $\displaystyle x=-3$ you should have obtained $\displaystyle \frac{0}{0}$ which is undefined. Whenever this happens, you know you need to perform some manipulations since the direct substitution property of limits only applies when the x-value is in the domain of the function. In general

$\displaystyle \lim_{x->a}f(x)=f(a)$ if $\displaystyle a$ is in $\displaystyle D_f$

3. I gotcha, thanks.

$\displaystyle \lim_{t \to 0} \frac{1}{t \sqrt{(1+t)}}-\frac{1}{t}$

I've simplified it to

$\displaystyle \lim_{t \to 0} \frac{1-\sqrt{(1+t)}}{t\sqrt{1+t}}$,

But tricks like rationalizing the numerator doesn't seem to work. Any ideas?

4. Hello, =rowe!

So what about this one: . $\displaystyle \lim_{t \to 0} \frac{1}{t \sqrt{1+t}}-\frac{1}{t}$

I've simplified it to: .$\displaystyle \lim_{t \to 0} \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$,

But tricks like rationalizing the numerator doesn't seem to work. . Yes, it does . . .
Multiply by $\displaystyle \frac{1+\sqrt{1+t}}{1+\sqrt{1+t}}$

. . $\displaystyle \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}\cdot\frac{1+\sqrt{1+t}}{1 +\sqrt{1+t}} \;=\;\frac{1 - (1+t)}{t\sqrt{1+t}\,(1+\sqrt{t+1})}$ .$\displaystyle =\;\frac{-t}{t\sqrt{1+t}\,(1+\sqrt{1+t})} \;=\;\frac{-1}{\sqrt{1+t}\,(1 + \sqrt{1+t})}$

Then: .$\displaystyle \lim_{t\to0}\frac{-1}{\sqrt{1+t}\,(1 + \sqrt{1+t})} \;=\;\frac{-1}{\sqrt{1+0}\,(1 + \sqrt{1+0})} \;=\;\frac{-1}{(1)(2)} \;=\;-\frac{1}{2}$