Evaluate This Limit

Calculus Level 2

$\text{If } \hspace{.1cm} \lim_{x \to 0} \frac{ae^x-b\cos x +ce^{-x}}{x \sin x}= 2,$

then find $a+b +c$ .

The answer is 4.

3 solutions

Lukas Leibfried
Sep 21, 2015

$\lim _{ x\rightarrow 0 }{ \frac { a{ e }^{ x }-b\cos { x } +c{ e }^{ -x } }{ x\sin { x } } } =\lim _{ x\rightarrow 0 }{ \frac { a{ e }^{ x }+b\sin { x } -c{ e }^{ -x } }{ \sin { x } +x\cos { x } } } =\lim _{ x\rightarrow 0 }{ \frac { a{ e }^{ x }+b\cos { x } +c{ e }^{ -x } }{ 2\cos { x } -x\sin { x } } } =\frac { a+b+c }{ 2 }$

$\frac{a + b + c}{2} = 2 \therefore a + b + c = \boxed{4}$

you're only allowed to use the L'Hopital's Rule when a lim is of the shape of 0/0 or inf/inf. I might be crazy but I don't see how you can use it here since this is of the form a-b+c/0. Maybe I'm just not seeing something here

Filip Bašić - 5 years, 6 months ago

I see what you're saying. I guess I should have specified that we know the limit is of indeterminate form because the answer is a non-zero finite number, therefore L'Hopital's Rule can be used.

Lukas Leibfried - 5 years, 6 months ago

The problem of using L"Hopital's rule here is that: lim g'(x) =0 where g(x)=xsinx and x -->0 and as i remember you can use this rule only when lim g'(x) is not 0. Correct me please if i'm wrong.

Liviu Tudor - 5 years, 5 months ago

@Liviu Tudor – But you could still have a situation in which lim f'(x)=0, which would generate another indeterminate form, setting up for a second use of L'Hopital's rule. In this second use, the denominator is now nonzero with direct substitution.

Joshua Solomon - 5 years, 5 months ago

Sri Harsha Koribille
Dec 25, 2015

using series expansions,

Noel Lo
Dec 22, 2015

Creative problem! Love it!!

Evaluate This Limit

The answer is 4.

3 solutions

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