Lt(x to 5) (|x-5|)/(x-5)=...

`Lt_(x to 5) (|x-5|)/(x-5)=`

A

0

B

1

C

`-1`

D

does not exist

Text Solution

Verified by Experts

The correct Answer is:

D

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AAKASH SERIES-LIMITS-EXERCISE-I

underset(x to -∞ )"Lt" (2|x|+x)/(5|x|-3x)=
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Lt(x to oo) ([x])/(x) (where[.] denotes greatest integer function )=
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If a gt 0, Lt(x to oo) ([ax+b])/(x)=
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underset(x to oo)lim (|3x^(2)+1|)/(2x^(2)+1)=
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underset(x to 1)lim {(1)/(x-1)-(2)/(x^(2)-1)}=
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underset(x to 1//2)lim {(8x-3)/(2x-1)-(4x^(2)+1)/(4x^(2)-1)}=
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Lt(x to 0) ("cosec x"-(1)/(x))=
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underset(x to oo)lim {sqrt(x^(2)+ax+b)-x}=
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Lt(x to 0) (1+ax)^(1//bx)=
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Lt(x to oo) (1+(2)/(3x))^((7x+2)/(5))=
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underset(x to 0)"Lt" ((x+a)/(x+b))^(x+b)=
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Lt(x to 0) ((x-1+cosx)/(x))^(1//x)
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Lt(x to 1) (1+sin pi x)^(cot pi x)=
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Lt(x to (pi)/(2)) ((1+cosx)/(1-cosx))^(secx)=
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Lt(x to oo) (1+e^(-x))^(e^(x))=
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underset(x to 0)"Lt" (3|x|+x)/(7|x|-5x)=
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Lt(x to 5) (|x-5|)/(x-5)=
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Lt(x to 2+)(|x-2|)/(x-2)=
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Lt(x to pi)[x]=
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Lt(x to 2)[x]=
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