Limits and Derivatives

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New answer posted•a year ago
S
a year ago

15. limx→πsin (π−x)π (π−x)=1π×limx→πsin (π−x)π−x

=1π.

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New answer posted•a year ago
S
a year ago

14. limx→0sinaxsinbx=limx→0sinaxax×axsinbxbx×bx

=ab

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New answer posted•a year ago
S
a year ago

13. limx→0sinaxbx=1blimx→0sinaxax×a

=ablimx→0sinaxax 

=ab

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New answer posted•a year ago
S
a year ago

Kindly go through the solution

=12×(−2)=−14.

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New answer posted•a year ago
S
a year ago

11. limx→1ax2+bx+ccx2+bx+a=a+b+cc+b+a=1

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New answer posted•a year ago
S
a year ago

10.=limz→1[z13−113z−1]÷limz→1[z16−116z−1]

= 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,

limx→axn−anx−a=nan−1

So,

= 1 3 × 6 = 2

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New answer posted•a year ago
S
a year ago

Kindly go through the solution

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New answer posted•a year ago
S
a year ago

8.

=limx→3 (x+3) (x2+9)2x+1.

= (3+3) (32+9)2×3+1

=6×186+1

=1087

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New answer posted•a year ago
S
a year ago

7.

=limx→2x (3x+5)−2 (3x+5) (x−2) (x+2)

=limx→2 (x−2) (3x+5) (x−2) (x+2)

=limx→23x+5x+2

=3×2+52+2

=6+54

=114.

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New answer posted•a year ago
S
a year ago

6. limx→0(x+1)5−1x=limx→0x5+5x4+10x3+10x2+5x+1−1

Q6. limx→0(x+1)5−1x

6. limx→0(x+1)5−1x=limx→0x5+5x4+10x3+10x2+5x+1−1

= 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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