Solved on Mar 10, 2024

Find the limit of the function $-2 \log_{8}(3-x)$ as $x$ approaches 3 from the left.

STEP 1

Assumptions
1. We are asked to find the limit of the function $-2 \log _{8}(3-x)$ as $x$ approaches $3$ from the left.
2. The logarithm is base $8$ .
3. The function is continuous at all points in its domain.

STEP 2

Understand the behavior of the function as $x$ approaches $3$ from the left.
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x)$

STEP 3

Recognize that as $x$ approaches $3$ from the left, $(3-x)$ approaches $0$ from the right.

STEP 4

Since the logarithm function is undefined for non-positive values, we must ensure that $(3-x)$ remains positive as $x$ approaches $3$ from the left.

STEP 5

Rewrite the limit by substituting $u = 3-x$ .
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = \lim _{u \rightarrow 0+}-2 \log _{8}(u)$

STEP 6

Now, we can see that as $x$ approaches $3$ from the left, $u$ approaches $0$ from the right.

STEP 7

Evaluate the limit of the logarithmic function as $u$ approaches $0$ from the right.
$\lim _{u \rightarrow 0+}\log _{8}(u)$

STEP 8

Recall that $\log _{b}(x)$ approaches $-\infty$ as $x$ approaches $0$ from the right for any base $b > 1$ .

STEP 9

Since $8$ is greater than $1$ , the limit of $\log _{8}(u)$ as $u$ approaches $0$ from the right is $-\infty$ .
$\lim _{u \rightarrow 0+}\log _{8}(u) = -\infty$

STEP 10

Multiply the limit of the logarithmic function by $-2$ .
$\lim _{u \rightarrow 0+}-2 \log _{8}(u) = -2 \times (-\infty)$

STEP 11

Recognize that multiplying a negative number by negative infinity results in positive infinity.
$-2 \times (-\infty) = +\infty$

STEP 12

Conclude that the limit of the original function as $x$ approaches $3$ from the left is positive infinity.
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = +\infty$
The limit is $+\infty$ .

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Find the limit of the function $-2 \log_{8}(3-x)$ as $x$ approaches 3 from the left.

STEP 1

Assumptions
1. We are asked to find the limit of the function $-2 \log _{8}(3-x)$ as $x$ approaches $3$ from the left.
2. The logarithm is base $8$ .
3. The function is continuous at all points in its domain.

STEP 2

Understand the behavior of the function as $x$ approaches $3$ from the left.
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x)$

STEP 3

Recognize that as $x$ approaches $3$ from the left, $(3-x)$ approaches $0$ from the right.

STEP 4

Since the logarithm function is undefined for non-positive values, we must ensure that $(3-x)$ remains positive as $x$ approaches $3$ from the left.

STEP 5

Rewrite the limit by substituting $u = 3-x$ .
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = \lim _{u \rightarrow 0+}-2 \log _{8}(u)$

STEP 6

Now, we can see that as $x$ approaches $3$ from the left, $u$ approaches $0$ from the right.

STEP 7

Evaluate the limit of the logarithmic function as $u$ approaches $0$ from the right.
$\lim _{u \rightarrow 0+}\log _{8}(u)$

STEP 8

Recall that $\log _{b}(x)$ approaches $-\infty$ as $x$ approaches $0$ from the right for any base $b > 1$ .

STEP 9

Since $8$ is greater than $1$ , the limit of $\log _{8}(u)$ as $u$ approaches $0$ from the right is $-\infty$ .
$\lim _{u \rightarrow 0+}\log _{8}(u) = -\infty$

STEP 10

Multiply the limit of the logarithmic function by $-2$ .
$\lim _{u \rightarrow 0+}-2 \log _{8}(u) = -2 \times (-\infty)$

STEP 11

Recognize that multiplying a negative number by negative infinity results in positive infinity.
$-2 \times (-\infty) = +\infty$

STEP 12

Conclude that the limit of the original function as $x$ approaches $3$ from the left is positive infinity.
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = +\infty$
The limit is $+\infty$ .

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Find the limit of the function −2log⁡8(3−x)-2 \log_{8}(3-x)−2log8​(3−x) as xxx approaches 3 from the left.

STEP 1

Assumptions1. We are asked to find the limit of the function −2log⁡8(3−x)-2 \log _{8}(3-x)−2log8​(3−x) as xxx approaches 333 from the left.2. The logarithm is base 888.3. The function is continuous at all points in its domain.

STEP 2

Understand the behavior of the function as xxx approaches 333 from the left.lim⁡x→3−−2log⁡8(3−x)\lim _{x \rightarrow 3-}-2 \log _{8}(3-x)x→3−lim​−2log8​(3−x)

STEP 3

Recognize that as xxx approaches 333 from the left, (3−x)(3-x)(3−x) approaches 000 from the right.

STEP 4

Since the logarithm function is undefined for non-positive values, we must ensure that (3−x)(3-x)(3−x) remains positive as xxx approaches 333 from the left.

STEP 5

Rewrite the limit by substituting u=3−xu = 3-xu=3−x.lim⁡x→3−−2log⁡8(3−x)=lim⁡u→0+−2log⁡8(u)\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = \lim _{u \rightarrow 0+}-2 \log _{8}(u)x→3−lim​−2log8​(3−x)=u→0+lim​−2log8​(u)

STEP 6

Now, we can see that as xxx approaches 333 from the left, uuu approaches 000 from the right.

STEP 7

Evaluate the limit of the logarithmic function as uuu approaches 000 from the right.lim⁡u→0+log⁡8(u)\lim _{u \rightarrow 0+}\log _{8}(u)u→0+lim​log8​(u)

STEP 8

Recall that log⁡b(x)\log _{b}(x)logb​(x) approaches −∞-\infty−∞ as xxx approaches 000 from the right for any base b>1b > 1b>1.

STEP 9

Since 888 is greater than 111, the limit of log⁡8(u)\log _{8}(u)log8​(u) as uuu approaches 000 from the right is −∞-\infty−∞.lim⁡u→0+log⁡8(u)=−∞\lim _{u \rightarrow 0+}\log _{8}(u) = -\inftyu→0+lim​log8​(u)=−∞

STEP 10

Multiply the limit of the logarithmic function by −2-2−2.lim⁡u→0+−2log⁡8(u)=−2×(−∞)\lim _{u \rightarrow 0+}-2 \log _{8}(u) = -2 \times (-\infty)u→0+lim​−2log8​(u)=−2×(−∞)

STEP 11

Recognize that multiplying a negative number by negative infinity results in positive infinity.−2×(−∞)=+∞-2 \times (-\infty) = +\infty−2×(−∞)=+∞

STEP 12

Conclude that the limit of the original function as xxx approaches 333 from the left is positive infinity.lim⁡x→3−−2log⁡8(3−x)=+∞\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = +\inftyx→3−lim​−2log8​(3−x)=+∞The limit is +∞+\infty+∞.

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Find the limit of the function $-2 \log_{8}(3-x)$ as $x$ approaches 3 from the left.

Assumptions
1. We are asked to find the limit of the function $-2 \log _{8}(3-x)$ as $x$ approaches $3$ from the left.
2. The logarithm is base $8$ .
3. The function is continuous at all points in its domain.

Understand the behavior of the function as $x$ approaches $3$ from the left.
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x)$

Recognize that as $x$ approaches $3$ from the left, $(3-x)$ approaches $0$ from the right.

Since the logarithm function is undefined for non-positive values, we must ensure that $(3-x)$ remains positive as $x$ approaches $3$ from the left.

Rewrite the limit by substituting $u = 3-x$ .
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = \lim _{u \rightarrow 0+}-2 \log _{8}(u)$

Now, we can see that as $x$ approaches $3$ from the left, $u$ approaches $0$ from the right.

Evaluate the limit of the logarithmic function as $u$ approaches $0$ from the right.
$\lim _{u \rightarrow 0+}\log _{8}(u)$

Recall that $\log _{b}(x)$ approaches $-\infty$ as $x$ approaches $0$ from the right for any base $b > 1$ .

Since $8$ is greater than $1$ , the limit of $\log _{8}(u)$ as $u$ approaches $0$ from the right is $-\infty$ .
$\lim _{u \rightarrow 0+}\log _{8}(u) = -\infty$

Multiply the limit of the logarithmic function by $-2$ .
$\lim _{u \rightarrow 0+}-2 \log _{8}(u) = -2 \times (-\infty)$

Recognize that multiplying a negative number by negative infinity results in positive infinity.
$-2 \times (-\infty) = +\infty$

Conclude that the limit of the original function as $x$ approaches $3$ from the left is positive infinity.
$\lim _{x \rightarrow 3-}-2 \log _{8}(3-x) = +\infty$
The limit is $+\infty$ .