Find aa and bb so that function ff is continuous.

f(x)={sin(3x)(eax−1)2×3:x<0ax2+3x+b:0≤x<1tan2(x−1)(x−1)2:x≥1\begin{eqnarray*}
f(x)=\left\{\begin{array}{lcc}
\frac{\sin(3x)(e^{ax}-1)}{2x^3} &:&x<0 \\
ax^2+3x+b &:& 0\le x <1\\
\frac{\tan^2(x-1)}{(x-1)^2} &:&x\ge1
\end{array}
\right.
\end{eqnarray*}
Well, I've been thinking on this and no way except L'Hأ´pital's rule has come to my mind, and this way takes a while to solve. Is there any better way?
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Did you think about Taylor series ?
– Claude Leibovici
2 days ago
@ClaudeLeibovici Not yet
– Ali Seyfi
2 days ago
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2 Answers
2
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Hints:
limx→0−sin(3x)xeax−1x
\lim_{x\to0^-}\frac{\sin(3x)}{x}\frac{e^{ax}-1}{x}
is finite. But you still have another xx at the denominator, so the limit is infinite unlessâ€¦
limx→1+tan(x−1)x−1=limt→0+tantt=â€¦
\lim_{x\to1^+}\frac{\tan(x-1)}{x-1}=
\lim_{t\to0^+}\frac{\tan t}{t}=â€¦
I would say "unless.... a=0\;a=0\; ", but then also \;b=0\;\;b=0\; and ...etc .
– DonAntonio
2 days ago
a must be 0, that's ok but why b=0?
– Ali Seyfi
2 days ago
tan part's limit will be 1 so lim ax^2+3x+b lim ax^2+3x+b when x-->1 from left should be 1 too.

– Ali Seyfi

2 days ago

@AliSeyfi What’s f(0)f(0)?

– egreg

2 days ago

@egreg got it. tnx

– Ali Seyfi

2 days ago

Hints:

Knowing

\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\cos x\frac{\tan x}{x}=1,\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\cos x\frac{\tan x}{x}=1,

\lim_{x\to0}\frac{e^x-1}x=1,\lim_{x\to0}\frac{e^x-1}x=1,

L’Hospital is unnecessary.