Question

$\lim_{x\to\infty}\frac{d}{dx}\left(\frac{8x^2+bx+c}{4x+f}\right)=2m$, then $\lim_{x\to m^+}2^{\frac{1}{2(1-x)}}=$

Answer 1

0

Answer 2

First, we need to evaluate the derivative of the function $y = \frac{8x^2+bx+c}{4x+f}$ with respect to $x$.

Using the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$, where $u = 8x^2+bx+c$ and $v = 4x+f$.

$u' = 16x+b$ and $v' = 4$.

$\frac{d}{dx}\left(\frac{8x^2+bx+c}{4x+f}\right) = \frac{(16x+b)(4x+f) - (8x^2+bx+c)(4)}{(4x+f)^2}$

Expanding the numerator:

$(16x+b)(4x+f) - 4(8x^2+bx+c) = (64x^2 + 16xf + 4bx + bf) - (32x^2 + 4bx + 4c)$

$= 64x^2 + 16xf + 4bx + bf - 32x^2 - 4bx - 4c$

$= 32x^2 + 16xf + bf - 4c$

The derivative is:

$\frac{d}{dx}\left(\frac{8x^2+bx+c}{4x+f}\right) = \frac{32x^2 + 16xf + bf - 4c}{(4x+f)^2} = \frac{32x^2 + 16xf + bf - 4c}{16x^2 + 8xf + f^2}$

Next, we evaluate the limit of this derivative as $x\to\infty$.

$\lim_{x\to\infty}\frac{32x^2 + 16xf + bf - 4c}{16x^2 + 8xf + f^2}$

This is the limit of a rational function where the degree of the numerator and the denominator are equal (both are 2). The limit as $x\to\infty$ is the ratio of the leading coefficients.

$\lim_{x\to\infty}\frac{32x^2 + 16xf + bf - 4c}{16x^2 + 8xf + f^2} = \frac{32}{16} = 2$

The problem states that this limit is equal to $2m$.

$2m = 2$

Dividing by 2, we get $m=1$.

Now we need to evaluate the second limit: $\lim_{x\to m^+}2^{\frac{1}{2(1-x)}}$.

Substitute $m=1$:

$\lim_{x\to 1^+}2^{\frac{1}{2(1-x)}}$

Let's analyze the exponent $\frac{1}{2(1-x)}$ as $x \to 1^+$.

As $x \to 1$ from the right side ($x > 1$), the term $(1-x)$ approaches 0 from the negative side ($1-x < 0$).

Let $x = 1+h$, where $h \to 0^+$.

Then $1-x = 1-(1+h) = -h$.

The exponent becomes $\frac{1}{2(-h)} = -\frac{1}{2h}$.

As $h \to 0^+$, $2h \to 0^+$, so $\frac{1}{2h} \to +\infty$.

Therefore, $-\frac{1}{2h} \to -\infty$.

The limit becomes:

$\lim_{h\to 0^+} 2^{-\frac{1}{2h}} = 2^{-\infty}$

We know that $2^{-\infty} = \frac{1}{2^\infty} = \frac{1}{\infty} = 0$.

So, $\lim_{x\to 1^+}2^{\frac{1}{2(1-x)}} = 0$.