Solving Limits: How to Simplify \lim_{x->0}

In summary, the conversation is about solving a limit problem involving a complicated fraction with trigonometric functions. The person is struggling to simplify the expression and makes several typos before realizing their mistake and correcting it. The conversation ends with them thanking someone named Zaid for their help.
  • #1
Petrus
702
0
Hello MHB,
I am currently working with a old exam and it says.\(\displaystyle \lim_{x->0}\frac{3-e^{x^2}-2\cos(x)}{x^2\sin^2(x)}\)
and this is how far I got but struggle on the simplify
2zgf2xe.png

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:
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  • #2
Re: limit,taylor

I think you mean x goes to zero right ?
 
  • #3
Re: limit,taylor

ZaidAlyafey said:
I think you mean x goes to zero right ?
Yes ofcourse haha My bad..
 
  • #4
Re: limit,taylor

\(\displaystyle \lim_{x \to 0}\frac{(3-1-x^2-\frac{x^4}{2!}-\cdot \cdot \cdot ) -2+x^2-2 \cdot \frac{x^4}{4!} +\cdot \cdot \cdot }{x^3-\frac{x^5}{3!}+\cdot \cdot \cdot}\)

Sounds like the numerator dominates, unless you meant sin^2(x) in the denominator ..
 
  • #5
Re: limit,taylor

By the way , in the case of a limit we have to use the expansion of all functions near the point , which is approached by x , hence the functions must converge .
 
  • #6
Re: limit,taylor

ZaidAlyafey said:
\(\displaystyle \lim_{x \to 0}\frac{(3-1-x^2-\frac{x^4}{2!}-\cdot \cdot \cdot ) -2+x^2-2 \cdot \frac{x^4}{4!} +\cdot \cdot \cdot }{x^3-\frac{x^5}{3!}+\cdot \cdot \cdot}\)

Sounds like the numerator dominates, unless you meant sin^2(x) in the denominator ..
I am making so much typo..:( I did Edit it now!
 
  • #7
Re: limit,taylor

Petrus said:
I am making so much typo..:( I did Edit it now!

I am sure you can solve it yourself :)
 
  • #8
Re: limit,taylor

ZaidAlyafey said:
I am sure you can solve it yourself :)
Actually I believe you are correct! I did accidently had a sign wrong.. If someone is interested how to solve it. You can first cancel out all \(\displaystyle x^0, 3-1-2\) then you can also cancel out \(\displaystyle x^2, -x^2+\frac{2x^2}{2!}\) expand that square function at numerator and divide evrything by \(\displaystyle x^4\) and you get the answer \(\displaystyle - \frac{7}{12}\), what I did wrong is that I did not see the negative sign in bracket at top. I did take \(\displaystyle \frac{1}{2}-\frac{2}{4!}\) while it should be \(\displaystyle -\frac{1}{2}-\frac{2}{4!}\)
Thanks once again Zaid!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

FAQ: Solving Limits: How to Simplify \lim_{x->0}

What is a limit in calculus?

A limit in calculus represents the value that a function approaches as the input approaches a specific value. It is denoted by the symbol "lim" and is an essential concept in calculus for understanding the behavior of functions.

How do I solve a limit?

To solve a limit, you can use algebraic manipulations, substitution, or graphing to simplify the function and evaluate the limit. You may also need to use limit laws and other calculus techniques such as L'Hospital's rule to solve more complex limits.

What is the limit of a function at a specific point?

The limit of a function at a specific point is the value that the function approaches as the input (x-value) approaches that specific point. It is not necessarily the same as the value of the function at that point and can only be determined using limit notation.

How do I evaluate a limit at infinity?

To evaluate a limit at infinity, you can use the concept of one-sided limits and determine the behavior of the function as the input approaches positive or negative infinity. You may also need to use some calculus techniques, such as the squeeze theorem, to evaluate the limit at infinity.

What are some common techniques for solving limits?

Some common techniques for solving limits include direct substitution, factoring, rationalization, and trigonometric identities. Other techniques, such as the sandwich theorem and the intermediate value theorem, may be used for more challenging limits. It is also essential to have a good understanding of algebra and calculus concepts to solve limits effectively.

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