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

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Homework Help Overview

The problem involves evaluating the limit as \( x \) approaches 0 for the expression \( \frac{e^{\tan(x)} - e^x}{\tan(x) - x} \). The subject area pertains to calculus, specifically limits and the application of L'Hospital's Rule.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of L'Hospital's Rule and the resulting indeterminate forms encountered. There are attempts to substitute values and analyze derivatives, with some questioning the correctness of their steps.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to resolve the limit. Some have noted that repeated applications of L'Hospital's Rule continue to yield indeterminate forms, prompting inquiries into alternative methods.

Contextual Notes

There is mention of confusion regarding the terms in the denominator after applying L'Hospital's Rule, and participants are reflecting on potential mistakes in their calculations. The problem remains unsolved, but there is a sense of collaborative effort to clarify the reasoning involved.

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Homework Statement


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

Homework Equations


The Attempt at a Solution


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

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

I don't understand where am i wrong? :(
 
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Pranav-Arora said:

Homework Statement


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


Homework Equations





The Attempt at a Solution


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

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

I don't understand where am i wrong? :(
After applying L'Hopital's Rule, you're missing a term in the denominator.
[tex]\lim_{x→0} \frac{e^{tan(x)} \cdot sec^2(x)-e^x}{sec^2(x) - 1}[/tex]

This is still of the indeterminate form [0/0].
 
Mark44 said:
After applying L'Hopital's Rule, you're missing a term in the denominator.
[tex]\lim_{x→0} \frac{e^{tan(x)} \cdot sec^2(x)-e^x}{sec^2(x) - 1}[/tex]

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?
 
Pranav-Arora said:
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.
 
Dick said:
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. :)
 

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