Can't get out of a 00 indeterminate form for this limitlimx→0(2xtan⁡x)(1−ex)2

Set g(x)=2xtan⁡(x)h(x)=(1−ex)2, f(x)=g(x)h(x)Then as limx→0g(x)=limx→0h(x)=0 you have thatlimx→0g(x)h(x)=limx→0g′(x)h′(x)if the right hand side exists by l'Hôpital. But as limx→0g′(x)=limx→0h′(x)=0 you have thatlimx→0g′(x)h′(x)=limx→0g″(x)h″(x)if the right hand side exists by l'Hôpital. But by evaluating this you can see thatlimx→0g″(x)h″(x)=42=2And therefore you have your limit.

Question: "Can't get out of a 00 indeterminate form..."

High School
Calculus and Analysis
Calculus 1
Limits and continuity

arrostetzwyl

Answered question

2022-03-16

Can't get out of a

\frac{0}{0}

indeterminate form for this limit

lim_{x \to 0} \frac{(2 x \tan x)}{{(1 - e^{x})}^{2}}

Answer & Explanation

Jakayla Hayes

Beginner2022-03-17Added 7 answers

Set

g (x) = 2 x \tan (x) h (x) = {(1 - e^{x})}^{2}, f (x) = \frac{g (x)}{h (x)}

Then as

lim_{x \to 0} g (x) = lim_{x \to 0} h (x) = 0

you have that

lim_{x \to 0} \frac{g (x)}{h (x)} = lim_{x \to 0} \frac{g^{'} (x)}{h^{'} (x)}

if the right hand side exists by l'Hôpital. But as

lim_{x \to 0} g^{'} (x) = lim_{x \to 0} h^{'} (x) = 0

you have that

lim_{x \to 0} \frac{g^{'} (x)}{h^{'} (x)} = lim_{x \to 0} \frac{g^{″} (x)}{h^{″} (x)}

if the right hand side exists by l'Hôpital. But by evaluating this you can see that

lim_{x \to 0} \frac{g^{″} (x)}{h^{″} (x)} = \frac{4}{2} = 2

And therefore you have your limit.

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