Evaluating Limit Using L'Hospitals Rule $\tiny{205.q4.1}$

[karush]

$\tiny{205.q4.1}\\$
$\textsf{Evaluate the limit using L'Hospitals Rule }$
\begin{align}
\displaystyle
L_{q4}&=\lim_{{x}\to{1}} \frac{11x+11\sec(\pi x)}{4x-4{x}^{2}}
&L'H=\frac{f'}{g'}&= \frac{11+11\pi\sec(\pi x)\tan(\pi x)}{4-8x}
\\
&=\frac{11(1)+11\sec(\pi (1))}{4(1)+4(1)^2}
&&=\frac{11+11\pi\sec(\pi )\tan(\pi)}{4-8}
\\
&=\frac{11-11}{4-4}
&&=\frac{11+0}{-4}
\\
&=\frac{0}{0}
&&=-\frac{11}{4}
\end{align}
$\textsf{think this is ok, but sugestions??}$

 

[MarkFL]

What you did is fine, you could also write:

$$L=\lim_{x\to1}\left(\frac{11x+11\sec(\pi x)}{4x-4x^2}\right)=\frac{11}{4}\lim_{x\to1}\left(\frac{1}{x}\cdot\frac{x+\sec(\pi x}{1-x}\right)=\frac{11}{4}\lim_{x\to1}\left(\frac{1}{x}\right)\cdot\lim_{x\to1}\left(\frac{x+\sec(\pi x)}{1-x}\right)=\frac{11}{4}\lim_{x\to1}\left(\frac{x+\sec(\pi x)}{1-x}\right)$$

Now apply L'Hôpital's Rule:

$$L=\frac{11}{4}\lim_{x\to1}\left(\frac{1+\pi\sec(\pi x)\tan(\pi x)}{-1}\right)=-\frac{11}{4}\left(\frac{1+0}{1}\right)=-\frac{11}{4}$$

 

[topsquark]

What you did is fine, you could also write:

$$L=\lim_{x\to1}\left(\frac{11x+11\sec(\pi x)}{4x-4x^2}\right)=\frac{11}{4}\lim_{x\to1}\left(\frac{1}{x}\cdot\frac{x+\sec(\pi x}{1-x}\right)=\frac{11}{4}\lim_{x\to1}\left(\frac{1}{x}\right)\cdot\lim_{x\to1}\left(\frac{x+\sec(\pi x)}{1-x}\right)=\frac{11}{4}\lim_{x\to1}\left(\frac{x+\sec(\pi x)}{1-x}\right)$$

That first line screwed me up a bit! (Sweating) I didn't realize it went over to the next line.

-Dan

 

[karush]

always love the simlified version