Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
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Limit of 7*tan(9*x/5)/x
-
Limit of (2+x^2-3*x)/(3+x^2-4*x)
-
Limit of -2+e^x-e^(-x)-sin(x)
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Limit of (2-7*x+3*x^2)/(2-5*x+2*x^2)
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Identical expressions
- seven *tan(nine *x/ five)/x
- 7 multiply by tangent of (9 multiply by x divide by 5) divide by x
- seven multiply by tangent of (nine multiply by x divide by five) divide by x
- 7tan(9x/5)/x
- 7tan9x/5/x
- 7*tan(9*x divide by 5) divide by x
Limit of the function 7*tan(9*x/5)/x
The solution
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[src]
/ /9*x\\
|7*tan|---||
| \ 5 /|
lim |----------|
x->0+\ x /
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
Limit((7*tan((9*x)/5))/x, x, 0)
Detail solution
Let's take the limit
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
transform
=
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) \lim_{x \to 0^+} \frac{1}{\cos{\left(\frac{9 x}{5} \right)}} = \lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
Do replacement
$$u = \frac{9 x}{5}$$
then
$$\lim_{x \to 0^+}\left(\frac{7 \sin{\left(\frac{9 x}{5} \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{63 \sin{\left(u \right)}}{5 u}\right)$$
=
$$\frac{63 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{5}$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
The final answer:
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = \frac{63}{5}$$
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
transform
True
=
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) \lim_{x \to 0^+} \frac{1}{\cos{\left(\frac{9 x}{5} \right)}} = \lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
Do replacement
$$u = \frac{9 x}{5}$$
then
$$\lim_{x \to 0^+}\left(\frac{7 \sin{\left(\frac{9 x}{5} \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{63 \sin{\left(u \right)}}{5 u}\right)$$
=
$$\frac{63 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{5}$$
The limit
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
is first remarkable limit, is equal to 1.
The final answer:
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = \frac{63}{5}$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \tan{\left(\frac{9 x}{5} \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(\frac{x}{7}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(\frac{9 x}{5} \right)}}{\frac{d}{d x} \frac{x}{7}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{63 \tan^{2}{\left(\frac{9 x}{5} \right)}}{5} + \frac{63}{5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{63 \tan^{2}{\left(\frac{9 x}{5} \right)}}{5} + \frac{63}{5}\right)$$
=
$$\frac{63}{5}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 0^+} \tan{\left(\frac{9 x}{5} \right)} = 0$$
and limit for the denominator is
$$\lim_{x \to 0^+}\left(\frac{x}{7}\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \tan{\left(\frac{9 x}{5} \right)}}{\frac{d}{d x} \frac{x}{7}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{63 \tan^{2}{\left(\frac{9 x}{5} \right)}}{5} + \frac{63}{5}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{63 \tan^{2}{\left(\frac{9 x}{5} \right)}}{5} + \frac{63}{5}\right)$$
=
$$\frac{63}{5}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
One‐sided limits
[src]
/ /9*x\\
|7*tan|---||
| \ 5 /|
lim |----------|
x->0+\ x /
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
63/5
$$\frac{63}{5}$$
= 12.6
/ /9*x\\
|7*tan|---||
| \ 5 /|
lim |----------|
x->0-\ x /
$$\lim_{x \to 0^-}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
63/5
$$\frac{63}{5}$$
= 12.6
= 12.6
Other limits x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = \frac{63}{5}$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = \frac{63}{5}$$
$$\lim_{x \to \infty}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = 7 \tan{\left(\frac{9}{5} \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = 7 \tan{\left(\frac{9}{5} \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
More at x→-oo
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = \frac{63}{5}$$
$$\lim_{x \to \infty}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
More at x→oo
$$\lim_{x \to 1^-}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = 7 \tan{\left(\frac{9}{5} \right)}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right) = 7 \tan{\left(\frac{9}{5} \right)}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{7 \tan{\left(\frac{9 x}{5} \right)}}{x}\right)$$
More at x→-oo
The graph