Q. 53 - Calculus

Question

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)$

Step-by-Step Solution

Answer

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)=2$

Step 1. Given information

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)$

Step 2.Simplify

Since,

$\underset{x\to a}{\lim }\left[c·f\left(x\right)\right]=c·\underset{x\to a}{\lim }f\left(x\right)$

Therefore,

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)=2·\underset{x\to \hspace{0.17em}0}{\lim }\left(\frac{x}{e^x-1}\right)$

Step 3. Calculate the value

Apply L'Hopital's rule,

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)=2·\underset{x\to \hspace{0.17em}0}{\lim }\left(\frac{1}{e^x}\right)$

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)=2\frac{1}{e^0}$

$\underset{x\to 0}{\lim }\left(\frac{2x}{e^x-1}\right)=2$