Advance Calculus Limit question
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I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
$endgroup$
add a comment |
$begingroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
$endgroup$
add a comment |
$begingroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
$endgroup$
I'm trying to compute this limit without the use of L'Hopital's rule:
$$lim_{x to 0^{+}} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$
I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
edited 6 hours ago
Foobaz John
22.9k41552
22.9k41552
asked 6 hours ago
Kevin CalderonKevin Calderon
563
563
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3 Answers
3
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oldest
votes
$begingroup$
Write the limit as
$$
lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
$$
and use the fact that
$$
lim_{xto 0+}frac{-2}{x}=-infty.
$$
to find that the limit equals $-1$.
$endgroup$
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^{frac{1}{x}}$ and consider $y to +infty$
begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
& stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
& = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
& stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
end{eqnarray*}
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write the limit as
$$
lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
$$
and use the fact that
$$
lim_{xto 0+}frac{-2}{x}=-infty.
$$
to find that the limit equals $-1$.
$endgroup$
add a comment |
$begingroup$
Write the limit as
$$
lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
$$
and use the fact that
$$
lim_{xto 0+}frac{-2}{x}=-infty.
$$
to find that the limit equals $-1$.
$endgroup$
add a comment |
$begingroup$
Write the limit as
$$
lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
$$
and use the fact that
$$
lim_{xto 0+}frac{-2}{x}=-infty.
$$
to find that the limit equals $-1$.
$endgroup$
Write the limit as
$$
lim_{xto 0+}frac{1+4^{-2/x}}{-1+4^{-2/x}}
$$
and use the fact that
$$
lim_{xto 0+}frac{-2}{x}=-infty.
$$
to find that the limit equals $-1$.
answered 6 hours ago
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
add a comment |
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^{frac{1}{x}}$ and consider $y to +infty$
begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
& stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
& = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
& stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
end{eqnarray*}
$endgroup$
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^{frac{1}{x}}$ and consider $y to +infty$
begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
& stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
& = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
& stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
end{eqnarray*}
$endgroup$
add a comment |
$begingroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^{frac{1}{x}}$ and consider $y to +infty$
begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
& stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
& = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
& stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
end{eqnarray*}
$endgroup$
A substitution can be helpful, as it transforms the expression into a rational function:
- Set $y=4^{frac{1}{x}}$ and consider $y to +infty$
begin{eqnarray*} frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}
& stackrel{y=4^{frac{1}{x}}}{=} & frac{frac{1}{y}+y}{frac{1}{y}-y} \
& = & frac{frac{1}{y^2}+1}{frac{1}{y^2}-1} \
& stackrel{y to +infty}{longrightarrow} & frac{0+1}{0-1} = -1
end{eqnarray*}
answered 1 hour ago
trancelocationtrancelocation
13.5k1827
13.5k1827
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$begingroup$
$$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.
$endgroup$
add a comment |
$begingroup$
$$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.
$endgroup$
$$lim_{xto 0^+}dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=lim_{xto 0^+}dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$
Clearly as $xto 0^+$, $2/xto infty$. Since the power of $4$ is $-2/x$, it must go to $0$. Effectively we have $frac{0+1}{0-1}=-1$. Hence the required limit is $-1$.
answered 42 mins ago
Paras KhoslaParas Khosla
2,758423
2,758423
add a comment |
add a comment |
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