If both lim x - a f x and lim x - a g x and exist finitely and lim x - a g x -0- thenlim x - afx-gx-lim x - a fxx - a-lim x - a gxA- FalseB- True

The correct option is A False The given result is true for all the cases except when lim_x→ ag(x) is zero. So the result lim_x→ af(x)g(x) = lim_x→ af(x)lim_x→ ...

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Byju's Answer

Standard XI

Mathematics

Non Removable Discontinuities

If both lim x...

Question

If both $lim x \to a f (x)$ and $lim x \to a g (x)$ and exist finitely and $lim x \to a g (x) = 0$ , then
$lim x \to a \frac{f (x)}{g (x)} = \frac{lim x \to a f (x)}{lim x \to a g (x)}$

True

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False

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Solution

The correct option is B
False

The given result is true for all the cases except when $lim x \to a g (x)$ is zero.
So the result
$lim x \to a \frac{f (x)}{g (x)} = \frac{lim x \to a f (x)}{lim x \to a g (x)}$
is applicable only when $lim x \to a f (x)$ and $lim x \to a g (x)$ exist finitely, and $lim x \to a g (x) \neq 0$ .

Thus the given statement is wrong.

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If both limx→af(x) and limx→ag(x) and exist finitely and limx→ag(x)=0, then limx→af(x)g(x)=limx→af(x)limx→ag(x)

The correct option is B False The given result is true for all the cases except when limx→ag(x) is zero. So the result limx→af(x)g(x)=limx→af(x)limx→ag(x) is applicable only when limx→af(x) and limx→ag(x) exist finitely, and limx→ag(x)≠0. Thus the given statement is wrong.

If both $lim x \to a f (x)$ and $lim x \to a g (x)$ and exist finitely and $lim x \to a g (x) = 0$ , then
$lim x \to a \frac{f (x)}{g (x)} = \frac{lim x \to a f (x)}{lim x \to a g (x)}$