Let f(x)=loge⁡x2+exloge⁡x4+e2x. If limx→∞ f(x)=l and limx→−∞ f(x)=m, then

A
$l = m$
B
$l = 2 m$
C
$2 l = m$
D
$l + m = 0$

Solution:

We have,

$I = lim_{x \to \infty} f (x) = lim_{x \to \infty} \frac{\log (x^{2} + e^{x})}{\log_{e} (x^{4} + e^{2 x})} = lim_{x \to \infty} \frac{2 x + e^{x}}{x^{2} + e^{x}} \times \frac{x^{4} + e^{2 x}}{4 x^{3} + 2 e^{2 x}}$

$\Rightarrow l = lim_{x \to \infty} \frac{(2 \frac{x}{e^{x}} + 1) (\begin{matrix} x^{4} \\ e^{2 x} + 1 \end{matrix})}{(1 + \frac{x^{2}}{e^{x}}) (2 + \frac{4 x^{3}}{e^{2 x}})}$

$\Rightarrow l = \frac{(2 \times 0 + 1) (0 + 1)}{(1 + 0) (2 + 0)} = \frac{1}{2} [∵ lim_{x \to \infty} \frac{x^{n}}{e^{x}} = 0, n > 0]$

and, $m = lim_{x \to - \infty} \frac{\log_{e} (x^{2} + e^{x})}{\log_{e} (x^{4} + e^{2 x})} = lim_{x \to - \infty} \frac{2 x + e^{x}}{x^{2} + e^{x}} \times \frac{x^{4} + e^{2 x}}{4 x^{3} + 2 e^{2 x}}$

$\begin{array}{l} \Rightarrow m = lim_{x \to - \infty} \frac{(2 + \frac{e^{x}}{x}) (1 + \frac{e^{2 x}}{x^{4}})}{(4 + \frac{2 e^{2 x}}{x^{3}}) (1 + \frac{e^{x}}{x^{2}})} = \frac{(2 + 0) (1 + 0)}{(4 + 0) (1 + 0)} = \frac{1}{2} \\ ∴ m = l . \end{array}$

Let $f (x) = \frac{\log_{e} (x^{2} + e^{x})}{\log_{e} (x^{4} + e^{2 x})} .$ If $lim_{x \to \infty} f (x) = l$ and $lim_{x \to - \infty} f (x) = m,$ then

Solution:

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