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Here is a question I have:

I have to calculate the limit $ \displaystyle{ \lim_{t \to 0} \frac{e^t -1}{te^t} }$. Can we apply the L'Hopital rule or I have to write it as: $$ \displaystyle{ \lim_{t \to 0} \frac{e^t -1}{te^t} = \lim_{t \to 0} \frac{e^t-1}{t} \cdot \frac{1}{e^t} =1 \cdot 1=1 }$$

Is $ \displaystyle{ \lim_{t \to 0} \frac{e^t-1}{t} }$ a "basic" limit that cannot be calculated using L'Hopital rule?

Thank's in advance!

edit: I was made I typo. Now it is the correct.

Can we apply L'Hopital's rule to calculate the limit $ \displaystyle{ \lim_{t \to 0} \frac{e^t -1}{te^t} }$ ?

Sorry for the confusion.

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I think you got something wrong when writing the second limit. The $-1$ should be in the numerator, not denominator. –  Beni Bogosel Mar 27 '12 at 9:35
    
What you wrote in your third line is simply wrong. You can apply L'Hopital's rule for your original limit since it is in an indeterminate form 0/0 or $\infty/\infty$. You cannot apply the rule to the 2nd limit in your question since it is not in an indeterminate form. –  Lyapunov Mar 27 '12 at 9:36
    
Yes you are both right. I have edit it. –  passenger Mar 27 '12 at 9:37

4 Answers 4

up vote 2 down vote accepted

The initial limit you want to calculate is not used in proving that $(e^t)'=e^t$, so you can use l'Hospital. Either way, I think your argument is shorter.

Now I realize that you asked if you can calculate $\lim_{t \to 0} \frac{e^t -1}{t}$ using l'Hospital. You cannot do that. That limit is elementary, and you do not have to prove it every time. You cannot use l'Hospital for this limit because the limit itself is used when proving that $(e^t)'=e^t$.

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O.K for the limit $ \lim_{t \to 0} \frac{e^t-1}{t}$. But what about $\displaystyle{ \lim_{t \to 0} \frac{e^t-1}{te^t} }$ ? Can apply L'Hopital's rule here ? –  passenger Mar 27 '12 at 9:52
    
@passenger Yes. That is what I was saying in my first paragraph. –  Beni Bogosel Mar 27 '12 at 10:16
    
O.K I wanted to be sure. Thank you for your time! –  passenger Mar 27 '12 at 10:17

Your method is essentially equivalent to L'Hospital's rule for functions of this form, viz.

$$\rm \left(\frac{f(x)-f(0)}{x}\right) \left(\frac{1}{f(x)}\right)\ \to\ f'(0)\:\frac{1}{f(0)}$$

$$\rm \frac{(f(x)-f(0))'}{(x\:f(x))'}\ =\ \frac{f'(x)}{f(x)+x\:f'(x)}\ \to\ \frac{f'(0)}{f(0)} $$

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You can use Lopital,the only thing is that you have a mistake in the derivate of the denominator (+ and not -).You can simplify e^t. But the final limit is 1. My answer was for your original post.Now I see you changed the post.

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But I didn't derive. –  passenger Mar 27 '12 at 9:42
    
OK, you are write.So aply lopital. (-:Did you change the question?I think I sow something different. –  alpha.Debi Mar 27 '12 at 9:43

Observe that when $t$ approaches $0$ we have an indeterminate form that is:

$$\lim_{t\to 0}\frac{e^t-1}{te^t}=\frac{1-1}{0}=\frac{0}{0}$$ So we can apply the L'Hopital rule:(take the derivative of numerator and denominator w.r.t $t$) $$\lim_{t\to 0}\frac{e^t-1}{te^t}=\lim_{t\to 0}\frac{e^t}{te^t+e^t}=\frac{1}{0+1}=1.$$

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