Finding: \(\displaystyle\lim_{{{x}\to-\infty}}{\frac{{{6}{x}^{{2}}+{5}{\cos{{\left\lbrace\pi{x}\right\rbrace}}}}}{{\sqrt{{{x}^{{4}}+{\sin{{\left\lbrace{5}\pi{x}\right\rbrace}}}}}}}}\) I've tried using l'Hospitals rule, however

Briley Cabrera

Briley Cabrera

Answered question

2022-04-06

Finding: limx6x2+5cos{πx}x4+sin{5πx}
I've tried using l'Hospitals rule, however we will alway keep the cos(πx) expression, as well for sin(5πx) . Also, terms do not cancel out with sin and π since the product with 5

Answer & Explanation

titemomo8gjz

titemomo8gjz

Beginner2022-04-07Added 10 answers

Hint: The functions sine and cosine are bounded. Therefore, even though they can oscillate wildly at infinity, they can be controlled.
This implies that in limit operations, we have the following rules of thumb that are correct if used appropriately:
1.limx(0×sin(x))=0
2.limx(0×cos(x))=0
3.limx(sin(x))=0
4.limx(cos(x))=0
In all of the above equalities, 0 in the parentheses is meant to be taken as something infinitesimal. Note that 3. and 4. can be thought as special cases of 1. and 2. respectively.
Now factor out x2 in the numerator and x4 under the square root in the denominator and proceed.
If that hint is not enough, hover your mouse over the orange area:
limx6x2+5cosπxx4+5sin5πx=limxx2(6+5cosπxx2)x4(1+5sin5πxx4)=limx6+5cosπxx21+5sin5πxx4=6
Coupewopmergorlpt

Coupewopmergorlpt

Beginner2022-04-08Added 13 answers

Easy using Squeeze theorem
6x256x2+5cos{πx}6x2+5
and
6x416x2+sin{5πx}6x4+1
then for x<-1 we have
6x25x4+16x2+5cosπxx4+sin5πx6x2+5x41
By squeeze theorem we get
limx6x2+5cos{πx}x4+sin{5πx}=6

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?