0

I have tried several ways to find the limit and reviewed my textbooks but I'm stumped. Plotting it on to Desmos reveals that it's 5 but I don't know how to find the answer algebraically. $$\lim_{x\to0} \frac{e^{5x}-1}{x}$$

Rodrigo de Azevedo
  • 1
Brucc
  • 117
  • 7
  • 3
    Are you familiar with L'Hopitals rule? The application of this quickly shows that the result is 5. – Oli D Jan 12 '22 at 14:00
  • @OliD first time i've heard of it, but i will look into it, thanks! – Brucc Jan 12 '22 at 14:01
  • 1
    Alternatively if you expand the exponential term as a Taylor series you might see why! – Oli D Jan 12 '22 at 14:04
  • 3
    Use the definition of derivative. – Robert Israel Jan 12 '22 at 14:05
  • 1
    There are multiple ways to solve it like using L'Hopital's rule or using Talylor series expansion for $e^{5x}$ , L'Hoptial probably the easiest way try it and see if you get the correct solution or not – Youssef Mohamed Jan 12 '22 at 14:07
  • 2
    Do you know what a derivative is? If so then this is simple, but I suspect you may be building up to derivatives at the moment. – Caedmon Jan 12 '22 at 14:10
  • Let $f(x)=e^{5x}$. What is $f'(0)$? And what is the definition of $f'(0)$??? – David C. Ullrich Jan 12 '22 at 14:13
  • Robert's comment shows that we do not even need L'Hospital or Taylor here, the definition of the derivate of $e^{5x}$ at $x_0=0$ is enough. – Peter Jan 12 '22 at 14:33

1 Answers1

1

An idea:

$$x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)\implies$$

$$ e^{5x}-1=(e^x-1)(e^{4x}+e^{3x}+e^{2x}+e^x+1)$$

and now use the basic limit

$$\lim_{x\to0}\frac{e^x-1}x=1$$

and thus you don't need to use L'Hospital rule.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • If the ”basic limit” $\lim_{x\to0}\frac{e^x-1}x=1$ is known then $\frac{e^{5x}-1}{x} = 5 \cdot \frac{e^{5x}-1}{5x} \to 5$ follows immediately, without the need for polynomial division. – Martin R Jan 12 '22 at 14:13
  • 1
    If you know that "basic limit" it's just a matter of change of variable $t = 5 x$. – Robert Israel Jan 12 '22 at 14:15
  • @MartinR Only if you already know that $$\lim_{x\to x_0} \frac{e^{f(x)}-1}{f(x)}=1$$ for any function $;f;$ such that $;f(x)\xrightarrow[x\to x_o]{}0;$ . I am trying to use the minimal ammount of stuff here. Same thing with calculating limits by means of variable changes... And the limit I called "basic" is usually given as such in elementary theory of limits, so it is up to the OP to choose what he can use. – DonAntonio Jan 12 '22 at 14:18