Learn L’Hospital’s rule function and evaluating L’hospital’s Rule by exponential, logarithmic or trigonometric function.
\(\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=0\) then
\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\)
If the limits on the right side exist.
\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\)
If \(\dfrac {0}{0}\) form is removed, the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.
A \(\dfrac {1}{4}\)
B \(\dfrac {1}{2}\)
C \(\dfrac {3}{2}\)
D \(\dfrac {3}{4}\)
A \(\lim\limits_{x\to \dfrac {\pi}{2}}\;\dfrac {1-sin\,x}{1+cos\,2x}\)
B \(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,2x}\)
C \(\lim\limits_{x\to 0}\;\dfrac {1-sin\,x}{1+cos\,x}\)
D \(\lim\limits_{x\to \pi}\;\dfrac {1-sin\,x}{1+sin\,x}\)
\(\lim\limits_{x\to a}\,f(x)= \lim\limits_{x\to a}\,g(x)=\infty\) then
\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\)
if the limits on the right side exist.
\(\lim\limits_{x\to a}\,\dfrac {f(x)}{g(x)}= \lim\limits_{x\to a}\,\dfrac {f'(x)}{g'(x)}\to\)If \(\Big(\dfrac {\infty}{\infty}\Big)\) form is removed the limits can be evaluated by direct substitution, otherwise we apply L' Hospital Rule again.
A \(\dfrac {1}{2}\)
B \(0\)
C \(\ell n\,2\)
D \(1\)
A \(\dfrac {2}{\pi}\)
B \(\dfrac {-2}{\pi}\)
C \(\dfrac {\pi}{2}\)
D \(\dfrac {-\pi}{2}\)
A \(\dfrac {1}{2\, \ell n\,2}\)
B \(2\,\ell n\,2\)
C \(\ell n\,2\)
D \(\dfrac {1}{\ell n\,2}\)
A \(-\pi\)
B \(-\dfrac {1}{\pi^2}\)
C \(\dfrac {1}{\pi^2}\)
D \(\pi\)
The Limits of the form \(\lim\limits_{x\to a}\;\Big(f(x)\Big)^{g(x)}\) can be of any one of the following indeterminate form
Each of the these limits can be evaluated by assuming the limits to be y and the taking log on both sides and applying L' Hospital rule.
A \(e^{(-2/\pi)}\)
B \(e^{(2/\pi)}\)
C \(e^{(\pi/2)}\)
D \(e^{(-\pi/2)}\)