Register

GATE CSE 1995 | Question: 7(B)

1,477 views

1 Answer

Best answer
12 votes
12 votes
$\displaystyle \lim_{x\rightarrow 0}\frac{\sin x}{x} = \displaystyle \lim_{x\rightarrow 0}\frac{\cos x}{1}$ (Applying L'Hôpital's rule since 0/0 form)

$\qquad =1.$
• edited by
User Avatar
Voting & Accepting an answer helps improve the community
Answer:
← PreviousNext →
← Previous in categoryNext in category →

Related questions

13 votes
13 votes
1 answer
1
Kathleen asked Oct 8, 2014
2,616 views
GATE CSE 1995 | Question: 25a
Find the minimum value of $3-4x+2x^2$.
Find the minimum value of $3-4x+2x^2$.
User Avatar
1 votes
1 votes
1 answer
2
Kathleen asked Oct 8, 2014
1,098 views
GATE CSE 1995 | Question: 2.18
The solution of differential equation $y''+3y'+2y=0$ is of the form $C_1e^x+C_2e^{2x}$ $C_1e^{-x}+C_2e^{3x}$ $C_1e^{-x}+C_2e^{-2x}$ $C_1e^{-2x}+C_2e^{-x}$
The solution of differential equation $y''+3y'+2y=0$ is of the form$C_1e^x+C_2e^{2x}$$C_1e^{-x}+C_2e^{3x}$$C_1e^{-x}+C_2e^{-2x}$$C_1e^{-2x}+C_2e^{-x}$
User Avatar
19 votes
19 votes
3 answers
3
Kathleen asked Oct 8, 2014
2,722 views
GATE CSE 1995 | Question: 2.8
If the cube roots of unity are $1, \omega$ and $\omega^2$, then the roots of the following equation are $(x-1)^3 +8 =0$ $-1, 1 + 2\omega, 1 + 2\omega^2$ $1, 1 - 2\omega, 1 - 2\omega^2$ $-1, 1 - 2\omega, 1 - 2\omega^2$ $-1, 1 + 2\omega, -1 + 2\omega^2$
If the cube roots of unity are $1, \omega$ and $\omega^2$, then the roots of the following equation are $$(x-1)^3 +8 =0$$$-1, 1 + 2\omega, 1 + 2\omega^2$ $1, 1 - 2\omeg...
User Avatar
26 votes
26 votes
3 answers
4
Kathleen asked Oct 8, 2014
5,655 views
GATE CSE 1995 | Question: 1.21
In the interval $[0, \pi]$ the equation $x=\cos x$ has No solution Exactly one solution Exactly two solutions An infinite number of solutions
In the interval $[0, \pi]$ the equation $x=\cos x$ has No solutionExactly one solutionExactly two solutionsAn infinite number of solutions
User Avatar
Link Copied!