Limits and Derivatives

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

15. limxπsin (πx)π (πx)=1π*limxπsin (πx)πx

=1π.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

14. limx0sinaxsinbx=limx0sinaxax*axsinbxbx*bx

=ab

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

13. limx0sinaxbx=1blimx0sinaxax*a

=ablimx0sinaxax 

=ab

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

=12*(2)=14.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

11. limx1ax2+bx+ccx2+bx+a=a+b+cc+b+a=1

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

10.=limz1[z13113z1]÷limz1[z16116z1]

= l i m z 1 [ z 1 3 1 1 3 z 1 ] ÷ l i m z 1 [ z 1 6 1 1 6 z 1 ]

We know that,

limxaxnanxa=nan1

So,

= 1 3 * 6 = 2

New answer posted

4 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

8.

=limx3 (x+3) (x2+9)2x+1.

= (3+3) (32+9)2*3+1

=6*186+1

=1087

New answer posted

4 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

7.

=limx2x (3x+5)2 (3x+5) (x2) (x+2)

=limx2 (x2) (3x+5) (x2) (x+2)

=limx23x+5x+2

=3*2+52+2

=6+54

=114.

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

6. limx0(x+1)51x=limx0x5+5x4+10x3+10x2+5x+11

Q6. limx0(x+1)51x

6. limx0(x+1)51x=limx0x5+5x4+10x3+10x2+5x+11

= l i m x 0 x 4 + 5 x 3 + 1 0 x 2 + 1 0 x + 5 .

(0)4 + 5(0)3 + 10(0)2 + 10(0) + 5

= 5.

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