Question

lim x-> 0^+ (1+sin7x)^8/x

Answer 1

If you mean

$\displaystyle \lim_{x\to0^+} \frac{(1+\sin(7x))^8}x$

then the limit doesn't exist, since 1 + sin(7x) approaches 1 while the denominator approaches 0.

However, if instead you mean

$\displaystyle \lim_{x\to0^+}(1+\sin(7x))^{8/x}$

rewrite the limand as

$(1+\sin(7x))^{8/x} = \exp\left(\ln(1+\sin(7x))^{8/x}\right) \\\\ = \exp\left(\dfrac{8\ln(1+\sin(7x))}x\right) \\\\ =\exp\left(\dfrac{\ln(1+\sin(7x))}x\right)^8$

The exponential function is continuous at 0, so we can pass the limit through it:

$\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(1+\sin(7x))}x\right)^8 = \exp\left(\lim_{x\to0^+}\dfrac{\ln(1+\sin(7x))}x\right)^8$

The remaining limit takes the indeterminate form 0/0, since ln(1 + sin(7x)) approaches ln(1) = 0, and so does x in the denominator. Apply L'Hopital's rule:

$\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{7\cos(7x)}{1+\sin(7x)}\right)^8 = \exp\left(7\right)^8 = \left(e^7\right)^8 = \boxed{e^{56}}$