Question

evaluate: limit x tends to 0 e power 3x-1 divided by x​

Answer 1

S O L U T I O N:

Given Limit:

$\displaystyle \rm =\lim_{x\to0}\bigg(\dfrac{e^{-x}-1}{x}\bigg)$

If we put x = 0, we get:

$\displaystyle \rm =\dfrac{e^{0}-1}{0}$

$\displaystyle \rm =\dfrac{1-1}{0}$

$\displaystyle \rm =\dfrac{0}{0}$

Which is an indeterminate form.

So, let us use L'Hopital's rule here.

Given Limit:

$\displaystyle \rm =\lim_{x\to0}\bigg(\dfrac{e^{-x}-1}{x}\bigg)$

Applying L'Hopital's rule, we get:

$\displaystyle \rm =\lim_{x\to0}\bigg(\dfrac{-e^{-x}}{1}\bigg)$

$\displaystyle \rm =\dfrac{-e^{0}}{1}$

$\displaystyle \rm =\dfrac{-1}{1}$

$\rm=-1$

Therefore:

$\displaystyle \rm\longrightarrow\lim_{x\to0}\bigg(\dfrac{e^{-x}-1}{x}\bigg)=-1$

⊕ Which is our required answer.

L E A R N  M O R E:

$\displaystyle \rm1. \: \: \lim_{x \to0} \sin(x) = 0$

$\displaystyle \rm2. \: \: \lim_{x \to0} \cos(x) = 1$

$\displaystyle \rm3. \: \: \lim_{x \to0} \dfrac{ \sin(x) }{x} = 1$

$\displaystyle \rm4. \: \: \lim_{x \to0} \dfrac{ \tan(x) }{x} = 1$

$\displaystyle \rm5. \: \: \lim_{x \to0} \dfrac{1 - \cos(x) }{x} = 1$

$\displaystyle \rm6. \: \: \lim_{x \to0} \dfrac{ \sin^{ - 1} (x) }{x} = 1$

$\displaystyle \rm7. \: \: \lim_{x \to0} \dfrac{ \tan^{ - 1} (x) }{x} = 1$

$\displaystyle \rm8. \: \: \lim_{x \to0} \dfrac{ {e}^{x} - 1 }{x} = 1$

$\displaystyle \rm9. \: \: \lim_{x \to0} \dfrac{ {a}^{x} - 1 }{x} = \ln(a)$

$\displaystyle \rm10. \: \: \lim_{x \to0} \dfrac{ \log(1 + x) }{x} = 1$

Answer 2

Answer:

-1

Step-by-step explanation:

Given limit,

$\longrightarrow \lim\limits_{x\to 0} \dfrac{e^{-x} - 1}{x}$

The given limit will attain (0/0) form by direct substitution. Therefore we have choose another method to evaluate our limit.

Consider again,

$\longrightarrow \lim\limits_{x\to 0} \dfrac{e^{-x} - 1}{x}$

$\longrightarrow \lim\limits_{x\to 0} \dfrac{ - (e^{-x} - 1)}{ - x}$

$\longrightarrow - \lim\limits_{ - x\to 0} \dfrac{e^{-x} - 1}{ - x}$

We have a standard limit:

• $\boxed{\lim\limits_{x\to 0} \dfrac{e^x-1}{x} = 1}$

Applying this will give us:

$\longrightarrow -(1)$

So the required answer is:

$\underline{\boxed{\lim\limits_{x\to 0} \dfrac{e^{-x} - 1}{x} = - 1}}$