Limits #1

$\begin{aligned} L & = \lim_{x \to \infty} \left(\frac {2x^2+3}{2x^2+5}\right)^{8x^2+3} & \small \color{#3D99F6} \text{A }1^\infty \text{ case, see note.} \\ & = \exp \left(\lim_{x \to \infty} (8x^2+3)\left(\frac {2x^2+3}{2x^2+5}-1\right)\right) \\ & = \exp \left(\lim_{x \to \infty} \frac {-2(8x^2+3)}{2x^2+5} \right) \\ & = \exp \left(\lim_{x \to \infty} \frac {-16x^2-6}{2x^2+5} \right) \\ & = e^{-8} \end{aligned}$

$\implies m = \boxed{-8}$

Note: For $\small \lim_{x \to a} \left(\frac {f(x)}{g(x)}\right)^{h(x)} = 1^\infty$ , we have ( see method 2 ):

$\small \lim_{x \to a} \left(\frac {f(x)}{g(x)}\right)^{h(x)} = \exp \left(\lim_{x \to a} h(x) \left(\frac {f(x)}{g(x)} - 1 \right) \right)$

Old method $\boxed{\text{1.-}}$ $\large{\displaystyle{\lim_{x \rightarrow \infty} \left(\dfrac{2x^2+3}{2x^2+5}\right)^{8x^2+3}}} = \lim_{x \rightarrow \infty} \left(\dfrac{2x^2+5}{2x^2+3}\right)^{-(8x^2+3)} =$ $= \lim_{x \rightarrow \infty} \left(1 +\dfrac{2}{2x^2+ 3}\right)^{-(8x^2+3)} = \lim_{x \rightarrow \infty} \left(1 +\frac{1}{\frac{2x^2+ 3}{2}}\right)^{(\frac{(2x^2 + 3)}{2}) \cdot \left(\frac{-(8x^2 +3)}{\frac{2x^2 + 3}{2}}\right)} = e^{-8}$ $\boxed{\text{2.-}}$ $\large{\displaystyle{\lim_{x \rightarrow \infty} \left(\dfrac{2x^2+3}{2x^2+5}\right)^{8x^2+3}}} = \lim_{x \rightarrow \infty} \left(\dfrac{2x^2+5}{2x^2+3}\right)^{-(8x^2+3)} =$ $= \lim_{x \rightarrow \infty} \left(1 +\dfrac{2}{2x^2+ 3}\right)^{-(8x^2+3)} = e^{\lim_{x \to \infty} -(8x^2 + 3) \cdot \dfrac{2}{2x^2+ 3}} = e^{-8}$ due to $\ln(1 + x) \sim x, \text{ as } x \to 0$

Set $u = 2x^2 + 5$ . Then the entire limit may be rewritten as $\lim_{u\to\infty} \left(\frac{u-2}{u}\right)^{4u-17} = \lim_{u\to\infty} \left(1 - \frac{2}{u}\right)^{4u} \cdot \left(1 - \frac{2}{u}\right)^{-17} = e^{-8} \cdot 1^{-17} = e^{-8}$

Firstly, note that $\ ( \frac{2x^2 +3}{2x^2 +5})^3 = ( \frac{2+3/x^2}{2+5/x^2})^3 \rightarrow 1 \quad as \ n \rightarrow \infty \quad (1)$

By application of Algebra Of Limits (a.k.a AOL) (quotient rule then product rule). So assuming the given limit exists we have, taking the exponential of the log of the given expression : (and from (1)) $\lim_{x \rightarrow \infty} \left(\dfrac{2x^2 +3}{2x^2 +5}\right)^{8x^2 +3} = \lim_{x \rightarrow \infty} exp \left(8x^2 (Ln(2x^2 +3) - Ln(2x^2 +5)\right)$ Letting $\ u = 2x^2$ and using continuity of $\ exp:(0, \infty) \rightarrow \mathbb{R}$ $\ = \lim_{u \rightarrow \infty} exp \left(4u (Ln(u +3) - Ln(u+5))\right) = exp \left( \lim_{u \rightarrow \infty} (4u (Ln(u +3) - Ln(u+5)) ) \right) \quad (2)$ Now $\lim_{u \rightarrow \infty} \frac{u^2}{(u+3)(u+5)}$ exists and equals 1 (by same method used to obtain (1)) so we can apply L'Hopital's Rule (and AOL) to (2) as follows: $\ = exp \left(4 \lim_{u \rightarrow \infty} \dfrac{\frac{d}{du} (Ln(u+3) - Ln(u+5)) }{\frac{d}{du}(1/u)} \right) = exp \left(4 \lim_{u \rightarrow \infty} \dfrac{\frac{1}{u+3} - \frac{1}{u+5} }{ - \frac{1}{u^2}} \right) = exp \left(-8 \lim_{u \rightarrow \infty} \frac{u^2}{(u+3)(u+5)} \right)$ $\ = \boxed{exp(-8)}$

(1-1/f(x))^f(x) >>>>>> as f(x) tends infinity, this tends to e^(-1)

Use this as a hint