L’Hospital’s Rule
Practice question:
Question No: 1 [ Maximum Mark: 7 ] (Do not use GDC)
Evaluate \( \operatorname{Lim}_{x \rightarrow 0}\left[\frac{\sin x}{x}\right]\) | |||
Answer : \( \lim _{x \rightarrow 0}\left[\frac{\sin x}{x}\right]=1\) | |||
. . |
.
.
Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)
Evaluate \( \lim _{x \rightarrow 0}\left[\frac{x+\sin 2 x}{x-\sin 2 x}\right]\) | |||
Answer : \( \lim _{x \rightarrow 0}\left[\frac{x+\sin 2 x}{x-\sin 2 x}\right]=-3\) | |||
. . |
.
.
Question No: 3 [ Maximum Mark: 7 ] (Do not use GDC)
Evaluate \( \operatorname{Lim}_{x \rightarrow 0}\left[\frac{\tan x}{x+x^2}\right] \) | |||
Answer : \(\lim _{x \rightarrow 0}\left[\frac{\tan x}{x+x^2}\right]=1 \) | |||
. . |
.
.
Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)
Evaluate \( \lim _{x \rightarrow 1}\left[\frac{1-x^2+2 x^2 \ln x}{1-\sin \frac{\pi x}{2}}\right]\) | |||
Answer : \( \operatorname{Lim}_{x \rightarrow 1}\left[\frac{1-x^2+2 x \ln x}{1-\sin \pi / 2 x}\right]=\frac{16}{\pi^2} \) | |||
. . |
.
.
Question No: 5 [ Maximum Mark: 7 ] (Do not use GDC)
Evaluate \(\lim _{x \rightarrow 0}(5 x \cdot 3 \cot x) \) | |||
Answer : \(\lim _{x \rightarrow 0}[5 x \cdot 3 \cot x]=15 \) | |||
. . |
.