It is given that lim x - 0ea x-b x-1-x2-2- Then the value of -a-b- isA- 2B- 3C- 4D- 1

The correct option is C 4lim_x→ 0e^ax bx 1x^2=2=00(form) Using L'Hospital's rule, we get nbsp; lim_x→ 0ae^ax b2x=a b0(form) For limit to exist a b→ 0 ⇒ a=b ...

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Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

It is given t...

Question

It is given that $lim x \to 0 \frac{e^{a x} - b x - 1}{x^{2}} = 2$ . Then the value of $| a | + | b |$ is

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Solution

The correct option is D $4$
$lim x \to 0 \frac{e^{a x} - b x - 1}{x^{2}} = 2 = \frac{0}{0} (form)$
Using L'Hospital's rule, we get
$lim x \to 0 \frac{a e^{a x} - b}{2 x} = \frac{a - b}{0} (form)$
For limit to exist -
$a - b \to 0 \Rightarrow a = b$
Now using L'Hospital's rule again, we get
$lim x \to 0 \frac{a^{2} e^{a x}}{2} = 2 \Rightarrow a^{2} = 4 \Rightarrow | a | = 2 = | b | \Rightarrow | a | + | b | = 4$

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It is given that limx→0eax−bx−1x2=2. Then the value of |a|+|b| is

The correct option is D 4limx→0eax−bx−1x2=2=00(form) Using L'Hospital's rule, we get limx→0aeax−b2x=a−b0(form) For limit to exist - a−b→0⇒a=b Now using L'Hospital's rule again, we get limx→0a2eax2=2⇒a2=4⇒|a|=2=|b|⇒|a|+|b|=4

It is given that $lim x \to 0 \frac{e^{a x} - b x - 1}{x^{2}} = 2$ . Then the value of $| a | + | b |$ is