Show that Lt(xto0+)(|x|)/x=1...

Show that
`Lt_(xto0+)(|x|)/x=1`

LIMITS
AAKASH SERIES|Exercise SOLVED EXAMPLE|36 Videos
LIMITS
AAKASH SERIES|Exercise EXERCISE-2.1 (VERY SHORT ANSWER QUESTIONS)|20 Videos
INVERSET TRIGONOMETRIC FUNCTIONS
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LOCUS
AAKASH SERIES|Exercise PRACTICE EXERCISE|37 Videos

Show that Lt(xto0+)(|x|)/x=1
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Lt(x to 0)(x^(2)-5x+7)/(2x^(2)+3x-1)=
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Lt(x to 0) (sqrt(4+x)-""^(3) sqrt(8+x))/(x)=
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Lt(x to 0)(sin(x^(0)))/(x)=
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Lt(x to 0) (cos5x-cos3x)/(x^(2))
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Lt(x to 0) (cosx-cos4x)/(cos5x-cos7x)=
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Lt(x to 0) (cos5x-cos3x)/((sin4x-sin3x))=
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Lt(x to 0) (2^(5x)-2^(7x))/(x)=
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Lt(x to 0) (5^(x)-3^(x))/(sin^(-1)x)=
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Lt(x to 0) (a^(sin2x)-b^(sinx))/(x)=
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Lt(x to2)(5^(-x)-0.04)/(x-2)=
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Evaluate Lt(x to 0) (e^(x) -1)/(sqrt(1 + x) - 1)
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If underset(x to o)"Lt" (log(3+x)-log(3-x))/(x)=k then k=
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Statement-I : If Lt(x to 0)f(x) exists, then Lt(x to a)f(x)=Lt(x to0)f...
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If f(x)=-sqrt(25-x^(2))", then " underset(x to 1)"Lt" (f(x)-f(1))/(x-1...
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f(x)=3x^(2)-4x+5,Lt(x to 1) (f(x)-f(1))/(x-1)=
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Lt(x to 0)(e^(alphax)-e^(betax))/(sin alphax- sin betax)=
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Lt(x to pi//2)(a^(cotx)-a^(cosx))/(cotx-cosx)=
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Lt(x to 0)((1+x^(2))^(1//3)-(1-2x)^(1//4))/(x+x^(2))
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Let f(a)=g(a)=k and the their nth derivatives f^(n)(a),g^(n)(a) exixst...
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Lim(x to 0)(x)/(""^(3)sqrt(x^(2)+3x+8)-""^(3) sqrt(x^(2)-5x+8))=
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