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
Hello, Bobby77!
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.
$\displaystyle \lim_{x\to\frac{\pi}{2}}\frac{\sin x}{x} \:= \:\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{1}\:=\:0$
To apply L'Hopital's Rule, the function must approach an indeterminate form:
. . $\displaystyle \frac{0}{0},\;\frac{\infty}{\infty},\;0^0,\;\hdots$
This limit exists: .$\displaystyle \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} $