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

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In summary, the conversation is about evaluating a limit using L'Hospital's Rule. The original attempt resulted in an indeterminate form, but after applying the rule again, the denominator became zero. The conversation then discusses the possibility of using another method, and concludes with the realization that the third derivative of the denominator is not zero, allowing for the limit to be evaluated.
  • #1
Saitama
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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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  • #2
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].
 
  • #3
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?
 
  • #4
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.
 
  • #5
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. :)
 

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

What is the definition of a limit in calculus?

A limit in calculus is the value that a function approaches as the input variable gets closer and closer to a specific value. It is a fundamental concept used to analyze the behavior of functions at specific points and is denoted by the symbol "lim".

How do you evaluate a limit using algebraic manipulation?

To evaluate a limit using algebraic manipulation, you can try to simplify the expression by factoring, cancelling common terms, or using algebraic identities. Once the expression is simplified, you can plug in the value that the input variable is approaching to find the limit.

What is the L'Hôpital's rule and how is it used to evaluate limits?

L'Hôpital's rule states that if the limit of a function is in an indeterminate form (such as 0/0 or ∞/∞), then it can be evaluated by taking the derivative of the numerator and denominator separately and then finding the limit of the resulting expression. It is a useful tool in evaluating limits of complex functions.

How can you evaluate the given limit, $\frac{e^{tan(x)}-e^x}{tan(x)-x}$?

In this case, you can use L'Hôpital's rule since the limit is in the indeterminate form of 0/0. Taking the derivative of the numerator and denominator results in $\frac{sec^2(x)e^{tan(x)}-e^x}{sec^2(x)-1}$. Plugging in the value of x that the input variable is approaching, you can find the final value of the limit.

What is the significance of evaluating limits in science and mathematics?

Evaluating limits is important in science and mathematics as it allows us to analyze the behavior of functions and make predictions about their values at specific points. It is used in various areas such as calculus, physics, and economics, and plays a crucial role in understanding and solving complex problems.

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