Question

using l hospitals rule find lim x tends to 0
e^x -1/x

Answer 1

Answer:

lim x tends to 0

e^x -1/x....

Answer 2

$\rm{\lim _{x\to \:0}\left(\frac{e^x-1}{x}\right)=\lim _{x\to \:0}\left(\frac{\left(e^x-1\right)^{'\:}}{\left(x\right)^{'\:}}\right)}$

$\longrightarrow\rm{\lim _{x\to \:0}\left(\frac{e^x-1}{x}\right)=\lim _{x\to \:0}\left(\frac{e^x}{1}\right)}$

$\longrightarrow\rm{\lim _{x\to \:0}\left(\frac{e^x-1}{x}\right)=\frac{e^0}{1} = 1/1 = 1}$