Evaluating the Limit: $\frac{e^{tan(x)}-e^x}{tan(x)-x}$

[Saitama]

Homework Statement

Evaluate the following limit:-
\lim_{x→0} \frac{e^{tan(x)}-e^x}{tan(x)-x}

Homework Equations

The Attempt at a Solution

Here are my attempts:-
I firstly apply L'Hospital rule, i get:-
\lim_{x→0} \frac{e^{tan(x)} \cdot sec^2(x)-e^x}{sec^2(x)}

If i directly substitute 0 here, i get the answer to be 0 but WolframAlpha says its 1.

I don't understand where am i wrong? :(

 

[Mark44]

Homework Statement

Evaluate the following limit:-
\lim_{x→0} \frac{e^{tan(x)}-e^x}{tan(x)-x}

Homework Equations

The Attempt at a Solution

Here are my attempts:-
I firstly apply L'Hospital rule, i get:-
\lim_{x→0} \frac{e^{tan(x)} \cdot sec^2(x)-e^x}{sec^2(x)}

If i directly substitute 0 here, i get the answer to be 0 but WolframAlpha says its 1.

I don't understand where am i wrong? :(

After applying L'Hopital's Rule, you're missing a term in the denominator.
\lim_{x→0} \frac{e^{tan(x)} \cdot sec^2(x)-e^x}{sec^2(x) - 1}

This is still of the indeterminate form [0/0].

 

[Saitama]

After applying L'Hopital's Rule, you're missing a term in the denominator.
\lim_{x→0} \frac{e^{tan(x)} \cdot sec^2(x)-e^x}{sec^2(x) - 1}

This is still of the indeterminate form [0/0].

Lol, sorry. :D

Yeah, it is still of 0/0 form but when i again apply L'Hosiptal, it is still of 0/0 form. :(

If i try to apply L'Hospital rule again and again, i am getting a 0 in denominator which will make my answer infinity, and the answer isn't that.

Is there any other method?

 

[Dick]

Lol, sorry. :D

Yeah, it is still of 0/0 form but when i again apply L'Hosiptal, it is still of 0/0 form. :(

If i try to apply L'Hospital rule again and again, i am getting a 0 in denominator which will make my answer infinity, and the answer isn't that.

Is there any other method?

The first and second derivatives the denominator are zero. The third derivative isn't.

 

[Saitama]

The first and second derivatives the denominator are zero. The third derivative isn't.

Yeah it isn't.

Sorry again. I was doing some oral calculation, so i think i made a mistake. Sorry. :)

EDIT: Thanks for all the help. I have solved the problem. :)