Solved Problems on Exponential Limits

First, recall the main formulas in the context of exponential/logarithmic limits. They will be widely used in the problems below:

Formula 1. $\lim\limits_{x \to 0} \dfrac{e^x-1}{x}=1$

Formula 2. $\lim\limits_{x \to 0} \dfrac{\log_e(1+x)}{x}=1$

Formula 3. $\lim\limits_{x \to 0} \dfrac{a^x-1}{x}=\log_e a$

Exponential Logarithmic Limits: Problems and Solutions

Problem 1:  $\lim\limits_{x \to 0} \dfrac{e^{ax}-1}{x}$


Put $z=ax.$ Then $z \to 0$ as $x \to 0$

Now $\lim\limits_{x \to 0} \dfrac{e^{ax}-1}{x}$ $=\lim\limits_{x \to 0} \dfrac{e^{ax}-1}{ax} \cdot a$ $=\lim\limits_{z \to 0} \dfrac{e^z-1}{z} \cdot \lim\limits_{x \to 0} a$

$=1 \cdot a $ [by Formula 1] $=a \quad$ ans.

Problem 2:  $\lim\limits_{x \to 0} \dfrac{e^{3x}-1}{\log(1+3x)}$


L = $\lim\limits_{x \to 0} \dfrac{e^{3x}-1}{\log(1+3x)}$ $=\lim\limits_{x \to 0} [\dfrac{e^{3x}-1}{3x} \cdot \dfrac{3x}{\log(1+3x)}]$

$=\lim\limits_{x \to 0} \dfrac{e^{3x}-1}{3x} \cdot \lim\limits_{x \to 0}\dfrac{3x}{\log(1+3x)}$

$=\lim\limits_{x \to 0} \dfrac{e^{3x}-1}{3x} \cdot\dfrac{1}{\lim\limits_{x \to 0}\dfrac{\log(1+3x)}{3x}}$

Put $z=3x$. So $z \to 0$ as $x \to 0$.

Therefore $L=\lim\limits_{z \to 0} \dfrac{e^{z}-1}{z} \cdot\dfrac{1}{\lim\limits_{z \to 0}\dfrac{\log(1+z)}{z}}$

$=1 \cdot \dfrac{1}{1}$ [by Formulas 1 & 2]

$ =1 \quad$ ans.

Spread the love
WhatsApp Group Join Now
Telegram Group Join Now