Answer to Question #129849 in Calculus for Hetisani Sewela

Question

Use L’Hopital’s rule to evaluate: ˆ

(a) limx→0 (1-cos^3x)/sin^2x

(b) limx→∞ (cos 1/x)^x

(c) limx→∞ x ln x/(x^2 - 1)

(d) limx→1 x ln x /(x^2 - 1)

Answer 1

a)applying L Hopital's rule

$\lim _{x\to \:0}\frac{3\cos ^2\left(x\right)\sin \left(x\right)}{2sinxcosx}$

$\lim _{x\to \:0}\frac{3\cos ^2\left(x\right)\sin \left(x\right)}{\sin \left(2x\right)}$

Again applying the rule ,

$\lim _{x\to \:0}\left(\frac{3\left(-\sin \left(2x\right)\sin \left(x\right)+\cos ^3\left(x\right)\right)}{\cos \left(2x\right)\cdot \:2}\right)$

Plugin the value for x=0,

$=\frac{3\left(-\sin \left(2\cdot \:0\right)\sin \left(0\right)+\cos ^3\left(0\right)\right)}{\cos \left(2\cdot \:0\right)\cdot \:2}$

= $\frac{3}{2}$

b) $\lim _{x\to \infty \:}\left(\cos ^x\left(\frac{1}{x}\right)\right)$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$

$\cos ^x\left(\frac{1}{x}\right)=e^{x\ln \left(\cos \left(\frac{1}{x}\right)\right)}$

$=\lim _{x\to \infty \:}\left(e^{x\ln \left(\cos \left(\frac{1}{x}\right)\right)}\right)$

$\mathrm{Apply\:the\:Limit\:Chain\:Rule}:\lim _{x\to \infty \:}\left(e^{\frac{\ln \left(\cos \left(\frac{1}{x}\right)\right)}{\frac{1}{x}}}\right)$

Applying L Hopitals's rule $\lim _{x\to \infty \:}\left(e^{\frac{\frac{\tan \left(\frac{1}{x}\right)}{x^2}}{-\frac{1}{x^2}}}\right)$

$\lim _{x\to \infty \:}e^{\left(-\tan \left(\frac{1}{x}\right)\right)}$

= $\lim _{x\to \infty \:}e^{\left(-\tan \left(\frac{1}{\infty}\right)\right)}$

= $\lim _{x\to \infty \:}e^{\left(-\tan (0)\right)}$

$\lim _{x\to \infty \:}e^{(0)}$

=1

Hence $\lim _{x\to \infty \:}\left(\cos ^x\left(\frac{1}{x}\right)\right)=1$

c)$\mathrm{Divide\:by\:highest\:denominator\:power:}\:\frac{\frac{\ln \left(x\right)}{x}}{1-\frac{1}{x^2}}$

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

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

$\lim _{x\to \infty \:}\left(\frac{\ln \left(x\right)}{x}\right)=0$ and ${\lim _{x\to \infty \:}\left(1-\frac{1}{x^2}\right)}=1$

=$\frac{0}{1}$ = 0

d)Applying L Hopital's rule

$=\lim _{x\to \:1}\left(\frac{\ln \left(x\right)+1}{2x}\right)$

$\mathrm{Plug\:in\:the\:value}\:x=1$

$=\frac{\ln \left(1\right)+1}{2\cdot \:1}$

$=\frac{1}{2}$