Question

Use L'Hospital to determine the following limit. Use exact values. \[ \lim _{x \rightarrow 0}(1+\sin 6 x)^{\frac{1}{x}}= \] What form is this limit? \( 0^{0} \) \( 1^{\infty} \) \( \infty^{0} \) \( \frac{0}{0} \) \( 0 \cdot \infty \) \( \frac{\infty}{\infty} \) \( 0^{\infty} \) Part 2 of 4 How should we rewrite the limit? \( \frac{1+\sin (6 x)}{x} \) \( e^{\ln (1+\sin (6 x))} \) \( \frac{x}{\frac{1}{1+\sin (6 x)}} \) \( e^{\frac{1}{x} \cdot \ln (1+\sin (6 x))} \) Now we focus on the limit of just the exponent, \( \frac{1}{x} \ln (1+\sin 6 x) \). \[ \lim _{x \rightarrow 0} \frac{1}{x} \ln (1+\sin (6 x))= \] \( \square \)

          Use L'Hospital to determine the following limit. Use exact values.
\[
\lim _{x \rightarrow 0}(1+\sin 6 x)^{\frac{1}{x}}=
\]

What form is this limit?
\( 0^{0} \)
\( 1^{\infty} \)
\( \infty^{0} \)
\( \frac{0}{0} \)
\( 0 \cdot \infty \)
\( \frac{\infty}{\infty} \)
\( 0^{\infty} \)
Part 2 of 4

How should we rewrite the limit?
\( \frac{1+\sin (6 x)}{x} \)
\( e^{\ln (1+\sin (6 x))} \)
\( \frac{x}{\frac{1}{1+\sin (6 x)}} \)
\( e^{\frac{1}{x} \cdot \ln (1+\sin (6 x))} \)

Now we focus on the limit of just the exponent, \( \frac{1}{x} \ln (1+\sin 6 x) \).
\[
\lim _{x \rightarrow 0} \frac{1}{x} \ln (1+\sin (6 x))=
\]
\( \square \)

Use L'Hospital to determine the following limit. Use exact values.

limx → 0(1+sin 6 x)^(1)/(x)=

What form is this limit?
0^0
1^∞
∞^0
(0)/(0)
0 ·∞
(∞)/(∞)
0^∞
Part 2 of 4

How should we rewrite the limit?
(1+sin (6 x))/(x)
e^ln (1+sin (6 x))
(x)/((1)/(1+sin (6 x)))
e^(1)/(x)·ln (1+sin (6 x))

Now we focus on the limit of just the exponent, (1)/(x)ln (1+sin 6 x).

limx → 0(1)/(x)ln (1+sin (6 x))=

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Added by Crystal A.

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