Given that f0-0 and lim x- 0 fx x exists- say L- Here f-0 denotes the derivative of f w-r-t x at x-0- Then L is

The correct option is D f'(0) Given, f(0)=0 and lim _ x→ 0 f(x) x #160; =L ∵ f'(0)=lim _ x→ 0 f(x) f(0) x 0 #160; =lim _ x→ 0 ...

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Standard VI

Mathematics

Cardinality

Given that ...

Question

Given that $f (0) = 0$ and $lim x \to 0 \frac{f (x)}{x}$ exists, say $L$ . Here $f^{'} (0)$ denotes the derivative of $f$ w.r.t $x$ at $x = 0$ . Then $L$ is

2 f^{'} (0) - 6

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2 f^{'} (0) - 5

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f^{'} (0)

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Solution

The correct option is D $f^{'} (0)$
Given, $f (0) = 0$ and $lim x \to 0 \frac{f (x)}{x} = L$
$∵ f^{'} (0) = lim x \to 0 \frac{f (x) - f (0)}{x - 0} = lim x \to 0 \frac{f (x)}{x}$
$∵ L = f^{'} (0)$

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Similar questions

Q. Given that

f (0) = 0

and

lim x \to 0 \frac{f (x)}{x}

exists, say

L

. Here

f^{'} (0)

denotes the derivative of

f

w.r.t.

x

x = 0

. Then

L

is :

Q. Let

f^{''} (x)

be continous at

x = 0

lim x \to 0 \frac{2 f (x) - 3 a f (2 x) + b f (8 x)}{{sin}^{2} x}

exists and

f (0) \neq 0, f^{'} (0) \neq 0

, then

Q. Let

f^{''} (x)

be continuous at

x = 0

.
If

lim x \to 0 \frac{2 f (x) - 3 a f (2 x) + b f (8 x)}{{sin}^{2} x}

exists and

f (0) \neq 0, f^{'} (0) \neq 0

then the value of

\frac{3 a}{b}

is ?

Q. If

f^{n} (x)

be continuous at

x = 0, f (0) \neq 0, f^{'} (0) \neq 0

and

lim x \to 0 \frac{2 f (x) - 3 a f (2 x) + b f (8 x)}{{sin}^{2} x}

exists. Then the values of

a

and

b

are

Q. If

f^{n} (x)

be continuous at

x = 0, f (0) \neq 0, f^{'} (0) \neq 0

and

lim x \to 0 \frac{2 f (x) - 3 a f (2 x) + b f (8 x)}{{sin}^{2} x}

,exists then values of

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Given that f(0)=0 and limx→0f(x)x exists, say L. Here f′(0) denotes the derivative of f w.r.t x at x=0. Then L is

The correct option is D f′(0)Given, f(0)=0 and limx→0f(x)x=L∵f′(0)=limx→0f(x)−f(0)x−0=limx→0f(x)x∵L=f′(0)

Given that $f (0) = 0$ and $lim x \to 0 \frac{f (x)}{x}$ exists, say $L$ . Here $f^{'} (0)$ denotes the derivative of $f$ w.r.t $x$ at $x = 0$ . Then $L$ is

The correct option is D $f^{'} (0)$
Given, $f (0) = 0$ and $lim x \to 0 \frac{f (x)}{x} = L$
$∵ f^{'} (0) = lim x \to 0 \frac{f (x) - f (0)}{x - 0} = lim x \to 0 \frac{f (x)}{x}$
$∵ L = f^{'} (0)$