Finding: limx→−∞6x2+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

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Finding: limx→−∞6x2+5cos⁡{πx}x4+sin⁡{5πx}I&#039;ve tried using l&#039;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

titemomo8gjz · Accepted Answer

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))=02.limx→∞(0×cos⁡(x))=03.limx→∞(sin⁡(x)∞)=04.limx→∞(cos⁡(x)∞)=0In 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:limx→−∞6x2+5cos⁡πxx4+5sin⁡5πx=limx→−∞x2(6+5cos⁡πxx2)x4(1+5sin⁡5πxx4)=limx→−∞6+5cos⁡πxx21+5sin⁡5πxx4=6

Coupewopmergorlpt · Accepted Answer

Easy using Squeeze theorem6x2−5≤6x2+5cos⁡{πx}≤6x2+5and6x4−1≤6x2+sin⁡{5πx}≤6x4+1then for x&amp;lt;-1 we have6x2−5x4+1≤6x2+5cos⁡πxx4+sin⁡5πx≤6x2+5x4−1By squeeze theorem we getlimx→−∞6x2+5cos⁡{πx}x4+sin⁡{5πx}=6

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

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