Question

limit x tends to 0 ((e^(x+3)-e^(3))/x )​

Answer 1

Answer:

use limits formula , take e^3 common out as it's a pure constant

Answer 2

Answer:

The value of the limit is $\lim_{x \to 0} \frac{e^{x+3}-e^{3}}{x}=e^{3}$

Step-by-step explanation:

The given limit is

$\lim_{x \to 0} \frac{e^{x+3}-e^{3}}{x}$

By substituting the value of the limit as x=0,we get

$\frac{e^{0+3}-e^{3}}{0} =\frac{0}{0}$

On substituting the limit we obtain an indeterminate form and hence we shall solve the limit using the L-hospital Rule in which the terms in the limit are differentiated till the limit gives a determinate form.By applying the rule,our limit becomes

$\lim_{x \to 0} \frac{\frac{d}{dx}( e^{x+3})-\frac{d}{dx} (e^{3})}{\frac{d}{dx} (x)}\\=\lim_{x \to 0}\frac{e^{x+3}-0}{1} =\lim_{x \to 0}e^{x+3}$

Substituting the value x=0 in the above limit,we get

$\lim_{x \to 0}e^{0+3}=e^{3}$

Therefore,The value of the limit is $\lim_{x \to 0} \frac{e^{x+3}-e^{3}}{x}=e^{3}$

#SPJ2