Question

Find the values of constant $a,b$ and $c$ so that
(i) $\underset{x\to 0}{lim}\frac{ax{e}^x-b\log (1+x)+cx{e}^{-x}}{x^2\sin x}=2$.

Answer 1

$\underset{x\to 0}{lim}\frac{ax{e}^x+cx{e}^x-\log (1+x)}{x^2\sin x}$

for $\underset{x\to 0}{lim}\sin x\to x$

and $\underset{x\to 0}{lim}\log (1+x)\to x$

So, we get,

$\underset{x\to 0}{lim}\frac{ax{e}^x+cx{e}^{-x}-bx}{x^2.x}$

$\underset{x\to 0}{lim}\frac{a{e}^x+c{e}^{-x}-b}{x^2}$

Now for this limit to exist,

$a+c-b=0$

$\Rightarrow a+c=b$.........(1)

Now we can apply t' Hospital

So we get,

$\underset{x\to 0}{lim}\frac{a{e}^x-c{e}^{-x}}{2x}$

For this limit to exist $a-c=0$......(2)

Now by L' Hospital we get

$\underset{x\to 0}{lim}\frac{a{e}^x+c{e}^x}{2}=2$

$\Rightarrow \frac{a+c}{2}=2$

$\Rightarrow a+c=4$........(3)

From (2) and (3),

$2a=4$

$\Rightarrow a=2$

and So, $c=2$

thus,

$\Rightarrow b=a+c$

$\Rightarrow b=2+2=4$

$\Rightarrow b=4$