First slide

Introduction to limits

Question

$f (x) = \frac{\ln (x^{2} + e^{x})}{\ln (x^{4} + e^{2 x})},$ Then $\lim_{x \to \infty} f (x)$ is equal to

Moderate

A

1
B

$\frac{1}{2}$
C

2
D

none of these

Solution

$\begin{array}{l} L = \lim_{x \to \infty} \frac{\ln (x^{2} + e^{x})}{\ln (x^{4} + e^{2 x})} = \lim_{x \to \infty} \frac{\ln e^{x} (1 + \frac{x^{2}}{e^{x}})}{\ln e^{2 x} (1 + \frac{x^{4}}{e^{2 x}})} \\ = \lim_{x \to \infty} \frac{x + \ln (1 + \frac{x^{2}}{e^{x}})}{2 x + \ln (1 + \frac{x^{4}}{e^{2 x}})} \\ = \lim_{x \to \infty} \frac{1 + \frac{1}{x} \ln (1 + \frac{x^{2}}{e^{x}})}{2 + \frac{1}{x} \ln (1 + \frac{x^{4}}{e^{2 x}})} \\ Note that as x \to \infty, \frac{x^{2}}{e^{x}} \to 0 and as x \to \infty, \frac{x^{2}}{e^{2 x}} \to 0 \\ (Using L' Hospital' s rule) \\ Hence L = \frac{1}{2} . \end{array}$

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