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Here , we can’t use L’ Hospital rule since limit is inifinity and what to do with (sinx)^2 ? Please someone explain.

in Calculus | 17 views

divide by the highest denominator power

$\lim _{x\to \infty \:}\left(\frac{x-\frac{\cos \left(x\right)}{x^2}}{1+\frac{\sin ^2\left(x\right)}{x^2}}\right)$

$=\frac{\lim _{x\to \infty \:}\left(x-\frac{\cos \left(x\right)}{x^2}\right)}{\lim _{x\to \infty \:}\left(1+\frac{\sin ^2\left(x\right)}{x^2}\right)}$

$=\frac{\infty \:}{1}$

$\mathrm{Apply\:Infinity\:Property:}\:\frac{\infty }{c}=\infty$

$=\infty \:$
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Does (sinx)^2 = sin^2x ?