X - 0limlog 1- 3x2xe5x - 1 -

The correct option is A 35 x → 0limlog(1 + 3x^2)x (e^5x 1) = x → 0limlog (1 + 3x^2)3x^2×5xe^5x 1×35 = 35x → 0limlog(1 + 3x^2)3x^2×x → 0 ...

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Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

x → 0limlog 1...

Question

$lim x \to 0 \frac{log (1 + 3 x^{2})}{x (e^{5 x} - 1)} =$

\frac{3}{5}

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\frac{5}{3}

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\frac{- 3}{5}

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\frac{- 5}{3}

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Solution

The correct option is A $\frac{3}{5}$
$lim x \to 0 \frac{log (1 + 3 x^{2})}{x (e^{5 x} - 1)}$
$= lim x \to 0 \frac{log (1 + 3 x^{2})}{3 x^{2}} \times \frac{5 x}{e^{5 x} - 1} \times \frac{3}{5}$
$= \frac{3}{5} lim x \to 0 \frac{log (1 + 3 x^{2})}{3 x^{2}} \times lim x \to 0 \frac{5 x}{e^{5 x} - 1}$
$= \frac{3}{5} \times 1 \times 1 = \frac{3}{5}$
$(∵ lim x \to 0 \frac{log (1 + x)}{x} = 1 and lim x \to 0 \frac{e^{x} - 1}{x} = 1)$

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Similar questions

Q. Select the expression you will get for x and y if you solve the following equations using the cross multiplication method.

\sqrt{3} x = 2

\frac{x}{3} = 7 - \frac{y}{5}

Add:

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Q. The set of the solutions for

\frac{(x + 1) (x - 3)}{(x + 5)} < 0

Q. The image of the line

\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}

in the plane

2 x - y + z + 3 = 0

is the line:

Find each of the following products.
(i) $(x - 4) (x - 4)$
(ii) $(2 x - 3 y) (2 x - 3 y)$
(iii) $(\frac{3}{4} x - \frac{5}{6} y) (\frac{3}{4} x - \frac{5}{6} y)$
(iv) $(x - \frac{3}{x}) (x - \frac{3}{x})$
(v) $(\frac{1}{3} x^{2} - 9) (\frac{1}{3} x^{2} - 9)$
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limx→0log(1+3x2)x(e5x−1)=

The correct option is A 35limx→0log(1+3x2)x(e5x−1)=limx→0log(1+3x2)3x2×5xe5x−1×35=35limx→0log(1+3x2)3x2×limx→05xe5x−1=35×1×1=35(∵limx→0log(1+x)x=1 and limx→0ex−1x=1)

$lim x \to 0 \frac{log (1 + 3 x^{2})}{x (e^{5 x} - 1)} =$