The value of lim x - 0ln 1-x-x is where x denotes the fractional part of x

The correct option is A Does not exist LHL = lim_x→ 0^ f(x) = lim_h→ 0 f(0 h) = lim_h→ 0In(1+{0 h})/{0 h} = lim_h→ 0In(2 h)/(1 h) = In 2 RHL = lim_x→ 0^+ f(x) ...

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Standard XII

Mathematics

Existence of Limit

The value of ...

Question

The value of $lim x \to 0 \frac{1 n (1 + {x})}{{x}}$ is (where {x} denotes the fractional part of x)

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Does not exist

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Solution

The correct option is D
Does not exist

LHL = $lim x \to 0^{-} f (x)$

= $lim h \to 0 f (0 - h)$

= $lim h \to 0 \frac{I n (1 + {0 - h})}{{0 - h}}$

= $lim h \to 0 \frac{I n (2 - h)}{(1 - h)}$ = $I n 2$

RHL = $lim x \to 0^{+} f (x)$ = $lim h \to 0 f (0 + h)$

= $lim h \to 0 \frac{I n (1 + {h})}{{h}}$ = $\frac{I n (1 + h)}{h}$ = 1

LHL $\neq$ RHL

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The value of limx→01n(1+{x}){x} is (where {x} denotes the fractional part of x)

The correct option is D Does not exist LHL = limx→0−f(x) = limh→0f(0−h) = limh→0In(1+{0−h}){0−h} = limh→0In(2−h)(1−h) = In 2 RHL = limx→0+f(x) = limh→0f(0+h) = limh→0In(1+{h}){h} = In(1+h)h = 1 LHL ≠ RHL

The value of $lim x \to 0 \frac{1 n (1 + {x})}{{x}}$ is (where {x} denotes the fractional part of x)