Question

16 mark for review which of the following limits does not yield an indeterminate form? a $lim_{x
ightarrow0}\frac{4x^{3}}{cos(x) - 1}$ b $lim_{x
ightarrow3}\frac{ln(x)}{x^{2}-7x + 12}$ c $lim_{x
ightarrowpi}\frac{pi - x}{sin(2x)-1}$ d $lim_{x
ightarrowinfty}\frac{x^{10}}{e^{2x}+x}$

Explanation:

Step1: Recall indeterminate forms

Indeterminate forms include $\frac{0}{0},\frac{\infty}{\infty},0\times\infty,\infty - \infty,0^{0},1^{\infty},\infty^{0}$.

Step2: Analyze option A

When $x
ightarrow0$, $\cos(x)-1
ightarrow\cos(0) - 1=1 - 1 = 0$ and $4x^{3}
ightarrow0$. So, $\lim_{x
ightarrow0}\frac{4x^{3}}{\cos(x)-1}$ is of the $\frac{0}{0}$ form.

Step3: Analyze option B

When $x
ightarrow3$, $x^{2}-7x + 12=(x - 3)(x - 4)
ightarrow(3 - 3)(3 - 4)=0$ and $\ln(\frac{x}{3})
ightarrow\ln(1)=0$. So, $\lim_{x
ightarrow3}\frac{\ln(\frac{x}{3})}{x^{2}-7x + 12}$ is of the $\frac{0}{0}$ form.

Step4: Analyze option C

When $x
ightarrow\pi$, $\sin(2x)-1=\sin(2\pi)-1=0 - 1=-1$ and $\pi - x
ightarrow0$. So, $\lim_{x
ightarrow\pi}\frac{\pi - x}{\sin(2x)-1}$ is of the $\frac{0}{-1}=0$ form, not an indeterminate form.

Step5: Analyze option D

When $x
ightarrow\infty$, $x^{10}
ightarrow\infty$ and $e^{2x}+x
ightarrow\infty$. So, $\lim_{x
ightarrow\infty}\frac{x^{10}}{e^{2x}+x}$ is of the $\frac{\infty}{\infty}$ form.

Answer:

C. $\lim_{x
ightarrow\pi}\frac{\pi - x}{\sin(2x)-1}$