Evaluate the limit

**Fenixx09** · September 1st 2011, 06:56 PM

$\\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t+\frac{a}{t}}{t+\frac{b}{t}} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t^{2}+at}{t^{2}+bt} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t(t+a)}{t(t+b)}$

The book gives the answer as $\frac{a}{b}$ but I can't figure out how to get to that solution. Up there is everything I have thought of and I'm stuck because it still has a zero in the denominator.

Can anyone give me a hint on what to do (no need to type any LaTeX code)

----------------------
Edit: I can't believe I made a mistake simply multiplying $(\frac{a}{t})(t) \& (\frac{b}{t})(t)$

**Prove It** · September 1st 2011, 08:01 PM

Originally Posted by Fenixx09

$\\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t+\frac{a}{t}}{t+\frac{b}{t}} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t^{2}+at}{t^{2}+bt} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t(t+a)}{t(t+b)}$

The book gives the answer as $\frac{a}{b}$ but I can't figure out how to get to that solution. Up there is everything I have thought of and I'm stuck because it still has a zero in the denominator.

Can anyone give me a hint on what to do (no need to type any LaTeX code)

----------------------
Edit: I can't believe I made a mistake simply multiplying $(\frac{a}{t})(t) \& (\frac{b}{t})(t)$

When you multiply top and bottom by $\displaystyle t$ you should actually get $\displaystyle \lim_{t \to 0}\frac{t^2 + a}{t^2 + b}$ .

So you have

$\displaystyle \begin{align*}\lim_{t \to 0} \frac{t^2 + a}{t^2 + b} &= \lim_{t \to 0}\frac{t^2 + b + a - b}{t^2 + b} \\ &= \lim_{t \to 0} \left(\frac{t^2 + b}{t^2 + b} + \frac{a - b}{t^2 + b}\right) \\ &= \lim_{t \to 0} \left(1 + \frac{a - b}{t^2 + b}\right) \\ &= 1 + \frac{a - b}{b} \\ &= 1 + \frac{a}{b} - \frac{b}{b} \\ &= 1 + \frac{a}{b} - 1 \\ &= \frac{a}{b} \end{align*}$

**CaptainBlack** · September 1st 2011, 08:30 PM

Originally Posted by Fenixx09

$\\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t+\frac{a}{t}}{t+\frac{b}{t}} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t^{2}+at}{t^{2}+bt} \\ \\ \\ \begin{array}{c}\lim\\t\rightarrow0\end{array} \ \ \frac{t(t+a)}{t(t+b)}$

The book gives the answer as $\frac{a}{b}$ but I can't figure out how to get to that solution. Up there is everything I have thought of and I'm stuck because it still has a zero in the denominator.

Can anyone give me a hint on what to do (no need to type any LaTeX code)

----------------------
Edit: I can't believe I made a mistake simply multiplying $(\frac{a}{t})(t) \& (\frac{b}{t})(t)$

$\lim_{t\to 0} \frac{t(t+a)}{t(t+b)} = \lim_{t\to 0} \frac{(t+a)}{(t+b)}=\frac{a}{b}$

CB

**Prove It** · September 1st 2011, 10:39 PM

Originally Posted by CaptainBlack

$\lim_{t\to 0} \frac{t(t+a)}{t(t+b)} = \lim_{t\to 0} \frac{(t+a)}{(t+b)}=\frac{a}{b}$

CB

That is true CB, but $\displaystyle \frac{t + \frac{a}{t}}{t + \frac{b}{t}} \neq \frac{t^2 + at}{t^2 + bt}$ ...

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