# limit of function.

• Nov 4th 2006, 10:38 AM
bobby77
limit of function.
An incorrect use of L hospital rule is illustrated in the following limit computation.Explain what is wrong and find correct value of limit.
lim x->pi/2 (sinx/x)= lim x->pi/2 (cosx/1)=0
• Nov 4th 2006, 10:42 AM
Plato
$\sin \left( {\frac{\pi }{2}} \right) = ?$
• Nov 4th 2006, 01:55 PM
Soroban
Hello, Bobby77!

Quote:

An incorrect use of L'Hopital's Rule is illustrated in the following limit computation.
Explain what is wrong and find correct value of limit.

$\lim_{x\to\frac{\pi}{2}}\frac{\sin x}{x} \:= \:\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{1}\:=\:0$

To apply L'Hopital's Rule, the function must approach an indeterminate form:
. . $\frac{0}{0},\;\frac{\infty}{\infty},\;0^0,\;\hdots$

This limit exists: . $\lim_{x\to\frac{\pi}{2}}\frac{\sin x}{x} \;=\;\frac{\sin\left(\frac{\pi}{2}\right)}{\frac{\ pi}{2}} \;=\;\frac{1}{\frac{\pi}{2}} \;=\;\frac{2}{\pi}$