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AAKASH SERIES-LIMITS-EXERCISE-I
- underset(x to oo)"Lt " 5^(x) sin (a/5^x)=
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- Lt(x to 2) (xf(2)-2f(x))/(x-2)=
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- underset(x to oo)lim (3x^(2)+5x+2)/(2x^(2)-3x+1)=
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- underset(x to oo)lim (2x+7)/(x^(2)+5x+4)=
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- underset(x to oo)lim (-x^(3)+8)/(2x^(2)+5x+7)=
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- If 7-(x^(2))/(12) le f(x) le7+(x^(3))/(5) for all n ne=0 then Lt(x to ...
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- Lt(x to oo) (x)/(sqrt(ax^(2)+b)-sqrt(cx^(2)+b))=
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- Lt(x to oo) ((n+2)!+(n+1)!)/((n+2)!-(n+1)!)=
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- Lt(x to oo) (3|x|+2x)/(4|x|-x)=
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- 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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