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A Major point which everyone have missed

LHospital Rule can be applied if

(lim x tends to a)f(x)/g(x) where f(x) approach 0 or infinity and g(x) approach 0 or infinity when x approach a respectively

or say we get (0/0) or (infinity/infinity) form

but the functions f(x) and g(x) should be continuous and differentiable at a

it is increasing as if you plot its graph it would be a sine curve around y=x line and whose keeps on incresing and decreasing but if you view function as a whole increases in its value hence it is increasing not strictly increasing.

f'(x)>0 for strictly increasing and f'(x) $\ge$ 0 for increasing

x=2tan^-1t

sinx=2t/(1+t²)

dx=2dt/(1+t²)

2 int dt/(t²+2t+1)

2 (-1/sinx+1)+C

$\mathop{\lim}\limits_{{x} \to {0}}(f(x)+g(x))$ ...... $\mathop{\lim}\limits_{{x} \to {0}}g(x)=\mathop{\lim}\limits_{{x} \to {0}}e^{\frac{log(\frac{1}{x})}{\frac{1}{sinx}}}=\frac{\rightarrow\infty}{\rightarrow\infty}form$ .........applying LH Rule........ $\mathop{\lim}\limits_{{x} \to {0}}e^ { x \frac{sin^2x}{x^2}\frac{1}{cosx}}$ ...... $\mathop{\lim}\limits_{{x} \to {0}}e^{(0)(1)(1)}=\mathop{\lim}\limits_{{x} \to {0}}e^0=1$ ........ $\mathop{\lim}\limits_{{x} \to {0}}f(x)$ ... $\mathop{\lim}\limits_{{x} \to {0}}(m(x))^{n(x)}=\mathop{\lim}\limits_{{x} \to {0}}e^{n(x)(m(x)-1)}$ ... $\mathop{\lim}\limits_{{x} \to {0}}e^{\frac{sinx-1}{x}}$ ...... $\mathop{\lim}\limits_{{x} \to {0}}e^{\rightarrow (-{\infty})}=0$ .......... $Hence\;answer\;is\;1$ ,,,,cheers

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yes because work done depends on initial and final posotion

hey rudra see this,it will answer all your queries

http://www.isid.ac.in/~rbb/crmosol_07.pdf