Find the limit of [(1+3x)^1/2 - (1-3x)^1/2]/x when x tends to 0.
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Lim(x→0) [(1+3x)½ - (1 - 3x)½]/x
check which form of limit exist here,
put x = 0, we see 0/0 is the form of limit here,
use the standard form
Lim(x→a) {xⁿ - aⁿ}/(x - a) = na^(n-1)
so,
Lim(x→0) [ {√(1+3x) - √1} - {√(1-3x) - √1}]/x
Lim(x→0) [1/2 × 3x - 1/2 (-3x) ]/x
Lim(x→0) [3x ]/x
= 3
check which form of limit exist here,
put x = 0, we see 0/0 is the form of limit here,
use the standard form
Lim(x→a) {xⁿ - aⁿ}/(x - a) = na^(n-1)
so,
Lim(x→0) [ {√(1+3x) - √1} - {√(1-3x) - √1}]/x
Lim(x→0) [1/2 × 3x - 1/2 (-3x) ]/x
Lim(x→0) [3x ]/x
= 3
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