1.3 Limits
Motivation ¶ What is the behavior of a function as its argument approaches a certain value?
For example, what happens to a function near boundaries of its domain? Near
“holes”? Under “extreme” cases, such as when something gets very small?
Limit (informal)
Let f ( x ) f(x) f ( x ) be defined on an open interval around x 0 x_0 x 0 , except possibly at x 0 x_0 x 0
itself. If f ( x ) f(x) f ( x ) is arbitrarily close to L L L for all x x x sufficiently close to
x 0 x_0 x 0 , we say f f f approaches the limit L L L as x x x approaches x 0 x_0 x 0 :
lim x → x 0 f ( x ) = L \lim_{x \to x_0} f(x) = L x → x 0 lim f ( x ) = L Some functions have limits. For example:
f ( x ) = x 2 − 1 x − 1 f(x) = \frac{x^2 - 1}{x - 1} f ( x ) = x − 1 x 2 − 1 is not defined when x = 1 x=1 x = 1 . What happens to it near 1? First, let’s factor and
simplify:
f ( x ) = ( x + 1 ) ( x − 1 ) x − 1 = x + 1 , x ≠ 1 f(x) = \frac{(x+1)(x-1)}{x-1} = x+1, \quad x \neq 1 f ( x ) = x − 1 ( x + 1 ) ( x − 1 ) = x + 1 , x = 1 So, we expect that f should be well-behaved and have a value that approaches
2 as x approaches 1! Let’s verify numerically:
x x x f f f 0.9 1.9 0.99 1.99 0.999 1.999 1.001 2.001 1.01 2.01 1.1 2.1
and also looking at a graph
However, some functions do not have limits. For example:
f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 is not defined at x = 0 x = 0 x = 0 . As x approaches zero from the positive or negative
side, the function “blows up” to ± ∞ \pm\infty ± ∞ . The sign depends on whether x
is positive or negative.
x x x f ( x ) f(x) f ( x ) 0.1 10 0.01 100 -0.01 -100 -0.1 -10
This divergence is clear on a plot:
We say that such a limit does not exist.
Limit (formal)
The limit of f f f as x x x approaches x 0 x_0 x 0 is L , denoted
lim x → x 0 f ( x ) = L \lim_{x \to x_0} f(x) = L x → x 0 lim f ( x ) = L if for every ϵ > 0 \epsilon > 0 ϵ > 0 there exists δ > 0 \delta > 0 δ > 0 such that for all x ,
if 0 < ∣ x − x 0 ∣ < δ 0 < |x-x_0| < \delta 0 < ∣ x − x 0 ∣ < δ then ∣ f ( x ) − L ∣ < ϵ |f(x)-L| < \epsilon ∣ f ( x ) − L ∣ < ϵ .
ϵ \epsilon ϵ is a tolerance setting how close f needs to be to L . Given
any ϵ \epsilon ϵ , we need to find a “window” of x values within δ \delta δ of
x 0 x_0 x 0 so that f stays within ϵ \epsilon ϵ of L . For example, consider the
following:
Based on this plot, if ϵ = 1 \epsilon = 1 ϵ = 1 , 3 < f < 5 3 < f < 5 3 < f < 5 if 3 < x < 5 3 < x < 5 3 < x < 5
(i.e., δ = 1 \delta = 1 δ = 1 ). For a general ϵ \epsilon ϵ ,
4 − ϵ < f < 4 + ϵ 4 - \epsilon < f < 4 + \epsilon 4 − ϵ < f < 4 + ϵ if 4 − δ < x < 4 + δ 4 - \delta < x < 4 + \delta 4 − δ < x < 4 + δ or
equivalently, ∣ x − 4 ∣ < ϵ |x-4| < \epsilon ∣ x − 4∣ < ϵ . We hence identify δ = ϵ \delta = \epsilon δ = ϵ and
have proven that
lim x → 4 f = 4 \lim_{x \to 4} f = 4 x → 4 lim f = 4 However, such a δ \delta δ cannot always be found for all ϵ \epsilon ϵ for certain
functions! There, the limit does not exist. For example, consider
f ( x ) = { x , x < 4 x − 2 , x ≥ 4 f(x) = \begin{cases}
x,& x < 4 \\
x-2,& x \ge 4
\end{cases} f ( x ) = { x , x − 2 , x < 4 x ≥ 4 Does the limit
lim x → 4 f ( x ) = 3 \lim_{x \to 4} f(x) = 3 x → 4 lim f ( x ) = 3 exist? If \epsilon = 0.5, there is no δ \delta δ that contains f around
x 0 = 4 x_0 = 4 x 0 = 4 . Therefore, this limit does not exsit
This math formalism is powerful, but clunky to apply. We will now move onto
using rules (limit laws) that have been proven using them!
Limit laws ¶ There are some simple limits and rules for combining them that are convenient
to know because they can be used to evaluate limits of more complicated
functions:
Taking limits ¶ “Nice” functions can be evaluated directly
lim x → − 1 x 3 + 4 x 2 − 3 = ( − 1 ) 3 + 4 ( − 1 ) 2 − 3 = − 1 + 4 − 3 = 0 \begin{align}
\lim_{x \to -1} x^3 + 4x^2 -3 &= (-1)^3 + 4(-1)^2 -3 \\
&= -1 + 4 -3 \\
&= 0
\end{align} x → − 1 lim x 3 + 4 x 2 − 3 = ( − 1 ) 3 + 4 ( − 1 ) 2 − 3 = − 1 + 4 − 3 = 0 This works even if they are complicated
lim x → 4 ( x 2 + 9 x + 1 ) 1 / 2 = ( 4 2 + 9 4 + 1 ) 1 / 2 = ( 25 5 ) 1 / 2 = 5 \begin{align}
\lim_{x \to 4} \left(\frac{x^2 + 9}{x + 1} \right)^{1/2}
&= \left(\frac{4^2 + 9}{4 + 1} \right)^{1/2} \\
&= \left(\frac{25}{5}\right)^{1/2} \\
&= \sqrt{5}
\end{align} x → 4 lim ( x + 1 x 2 + 9 ) 1/2 = ( 4 + 1 4 2 + 9 ) 1/2 = ( 5 25 ) 1/2 = 5 Limits are most useful when functions have “holes” (function is not defined at
x 0 x_0 x 0 )
lim x → 1 x 2 + x − 2 x 2 − x = lim x → 1 ( x + 2 ) ( x + 1 ) x ( x + 1 ) = lim x → 1 x + 2 x = 3 \begin{align}
\lim_{x \to 1} \frac{x^2 + x -2}{x^2 - x}
&= \lim_{x \to 1} \frac{(x+2)(x+1)}{x(x+1)} \\
&= \lim_{x \to 1} \frac{x+2}{x} \\
&= 3
\end{align} x → 1 lim x 2 − x x 2 + x − 2 = x → 1 lim x ( x + 1 ) ( x + 2 ) ( x + 1 ) = x → 1 lim x x + 2 = 3 Note that this function had a “hole” at x = 1 x = 1 x = 1 that we removed by factoring.
This works even if the factors are “ugly”:
lim x → 4 4 x − x 2 2 − x = lim x → 4 x ( 4 − x ) 2 − x = lim x → 4 x ( 2 + x ) ( 2 − x ) 2 − x = lim x → 4 x ( 2 + x ) = 4 ( 2 + 4 ) = 16 \begin{align}
\lim_{x \to 4} \frac{4x - x^2}{2 - \sqrt{x}}
&= \lim_{x \to 4} \frac{x(4 - x)}{2 - \sqrt{x}} \\
&= \lim_{x \to 4} \frac{x(2 + \sqrt{x})(2 - \sqrt{x})}{2 - \sqrt{x}} \\
&= \lim_{x \to 4} x(2 + \sqrt{x}) \\
&= 4(2 + \sqrt{4}) \\
&= 16
\end{align} x → 4 lim 2 − x 4 x − x 2 = x → 4 lim 2 − x x ( 4 − x ) = x → 4 lim 2 − x x ( 2 + x ) ( 2 − x ) = x → 4 lim x ( 2 + x ) = 4 ( 2 + 4 ) = 16 Here, we made use of the difference of squares identity:
a 2 − b 2 = ( a + b ) ( a − b ) a^2 - b^2 = (a + b)(a - b) a 2 − b 2 = ( a + b ) ( a − b ) Another useful trick is to multiply by a factor of “1” that makes a difference
of squares (multiply by conjugate)
lim x → 0 3 x + 1 − 1 x = lim x → 0 ( 3 x + 1 − 1 ) ( 3 x + 1 + 1 ) x ( 3 x + 1 + 1 ) = lim x → 0 ( 3 x + 1 ) 2 − 1 2 x ( 3 x + 1 + 1 ) = lim x → 0 3 x + 1 − 1 x ( 3 x + 1 + 1 ) = lim x → 0 3 3 x + 1 + 1 = 3 2 \begin{align}
\lim_{x \to 0} \frac{\sqrt{3x + 1} - 1}{x}
&= \lim_{x \to 0}
\frac{(\sqrt{3x + 1} - 1)(\sqrt{3x + 1} + 1)}{x(\sqrt{3x + 1} + 1)} \\
&= \lim_{x \to 0} \frac{(\sqrt{3x + 1})^2 - 1^2}{x(\sqrt{3x + 1} + 1)} \\
&= \lim_{x \to 0} \frac{3x + 1 - 1}{x(\sqrt{3x + 1} + 1)} \\
&= \lim_{x \to 0} \frac{3}{\sqrt{3x + 1} + 1} \\
&= \frac{3}{2}
\end{align} x → 0 lim x 3 x + 1 − 1 = x → 0 lim x ( 3 x + 1 + 1 ) ( 3 x + 1 − 1 ) ( 3 x + 1 + 1 ) = x → 0 lim x ( 3 x + 1 + 1 ) ( 3 x + 1 ) 2 − 1 2 = x → 0 lim x ( 3 x + 1 + 1 ) 3 x + 1 − 1 = x → 0 lim 3 x + 1 + 1 3 = 2 3 Limits at infinity ¶ How do functions behave when x gets “big”? For example, if the independent
variable is time, the behavior after waiting a long time may be a “steady state”.
Population dynamics
This may represent, for example, a population with limited resources.
Proportional-integral control
This may represent, for example, the response of a servo or valve over time.
Limit at infinity
lim x → ∞ f ( x ) = L \lim_{x \to \infty} f(x) = L x → ∞ lim f ( x ) = L if for every ϵ > 0 \epsilon > 0 ϵ > 0 , there exists a value M such that for all x ,
if x > M x > M x > M , ∣ f ( x ) − L ∣ < ϵ |f(x) - L| < \epsilon ∣ f ( x ) − L ∣ < ϵ .
i.e., f gets close to L when x is “big”.
Some important limits at infinity:
lim x → ± ∞ k = k \displaystyle\lim_{x \to \pm\infty} k = k x → ± ∞ lim k = k
lim x → ± ∞ 1 x = 0 \displaystyle\lim_{x \to \pm\infty} \dfrac{1}{x} = 0 x → ± ∞ lim x 1 = 0
lim x → − ∞ e x = 0 \displaystyle\lim_{x \to -\infty} e^x = 0 x → − ∞ lim e x = 0
Other limit laws (sum, quotient, etc.) also hold at infinity! If function is
already in a convenient form, use known limits to evaluate unknown ones. For
example,
lim x → ∞ 5 + 1 x = lim x → ∞ 5 + lim x → ∞ 1 x = 5 + 0 = 5 \lim_{x \to \infty} 5 + \frac{1}{x} =
\lim_{x \to \infty} 5 + \lim_{x \to \infty} \frac{1}{x} =
5 + 0 = 5 x → ∞ lim 5 + x 1 = x → ∞ lim 5 + x → ∞ lim x 1 = 5 + 0 = 5 For rational functions f ( x ) = p ( x ) / q ( x ) f(x) = p(x) / q(x) f ( x ) = p ( x ) / q ( x ) , divide by largest power of x in
denominator:
lim x → − ∞ 11 x + 2 2 x 3 − 1 = lim x → − ∞ 11 x x 3 + 2 x 3 2 x 3 x 3 + 1 x 3 = 0 + 0 2 − 0 = 0 \lim_{x \to -\infty} \frac{11x + 2}{2x^3-1} =
\lim_{x \to -\infty} \frac{ \dfrac{11x}{x^3} + \dfrac{2}{x^3} }{ \dfrac{2x^3}{x^3} + \dfrac{1}{x^3}}
= \frac{0 + 0}{2 - 0} = 0 x → − ∞ lim 2 x 3 − 1 11 x + 2 = x → − ∞ lim x 3 2 x 3 + x 3 1 x 3 11 x + x 3 2 = 2 − 0 0 + 0 = 0 lim x → ∞ 5 x 2 + 8 x − 3 3 x 2 + 4 = lim x → ∞ 5 + 8 x − 3 x 2 3 + 2 x 2 = 5 + 0 − 0 3 − 0 = 5 3 \lim_{x \to \infty} \frac{5x^2+8x-3}{3x^2+4} =
\lim_{x \to \infty} \frac{ 5 + \dfrac{8}{x} - \dfrac{3}{x^2} }{ 3 + \dfrac{2}{x^2} } =
\frac{5+0-0}{3 - 0} = \frac{5}{3} x → ∞ lim 3 x 2 + 4 5 x 2 + 8 x − 3 = x → ∞ lim 3 + x 2 2 5 + x 8 − x 2 3 = 3 − 0 5 + 0 − 0 = 3 5 Not all functions have limits at infinity. Some “blow up”!
lim x → ∞ 2 x 2 − 3 7 x + 4 = lim x → ∞ 2 x − 3 x 7 + 4 x \lim_{x \to \infty} \frac{2x^2-3}{7x+4} = \lim_{x \to \infty} \frac{ 2x - \frac{3}{x} }{ 7 + \frac{4}{x} } x → ∞ lim 7 x + 4 2 x 2 − 3 = x → ∞ lim 7 + x 4 2 x − x 3 This limit does not exist. A helpful shortcut for rational polynomials: check
coefficients of highest powers in numerator and denominator!
One-sided limits ¶ How do functions behave as you approach a point x 0 x_0 x 0 from a direction? One
physical example of this is an old question in statistical mechanics: when you
have vapor-liquid phase coexistence, does the density change stepwise or
continuously at the interface?
If the density as a function of distance from the interface is ρ ( z ) \rho(z) ρ ( z ) ,
the liquid density can be denoted as lim z → 0 − ρ ( z ) \lim_{z \to 0^-} \rho(z) lim z → 0 − ρ ( z ) , while the
vapor density can be denoted as lim z → 0 + ρ ( z ) \lim_{z \to 0^+} \rho(z) lim z → 0 + ρ ( z ) .
One-sided limit
If f is defined on the interval x 0 < x < b x_0 < x < b x 0 < x < b and f ( x ) f(x) f ( x ) approaches
arbitrarily close to L L L as x x x approaches x 0 x_0 x 0 , then f f f has a right-hand
limit L L L at x 0 x_0 x 0 :
lim x → x 0 + f ( x ) = L \lim_{x \to x_0^+} f(x) = L x → x 0 + lim f ( x ) = L (This limit can also be written more formally in δ \delta δ -ϵ \epsilon ϵ form.)
The same can be done for the left-hand limit if a < x < x 0 a < x < x_0 a < x < x 0
lim x → x 0 − f ( x ) = L \lim_{x \to x_0^-} f(x) = L x → x 0 − lim f ( x ) = L The two-sided limit exists if and only if both one-sided limits exist and have
the same value:
lim x → x 0 f ( x ) = L ⇔ lim x → x 0 + f ( x ) = lim x → x 0 − f ( x ) = L \lim_{x \to x_0} f(x) = L \Leftrightarrow
\lim_{x \to x_0^+} f(x) = \lim_{x \to x_0^-} f(x) = L x → x 0 lim f ( x ) = L ⇔ x → x 0 + lim f ( x ) = x → x 0 − lim f ( x ) = L For example, if
f ( x ) = ∣ x ∣ x = { − 1 , x < 0 1 , x > 0 f(x) = \frac{|x|}{x} = \begin{cases}
-1, & x<0\\
1, & x>0
\end{cases} f ( x ) = x ∣ x ∣ = { − 1 , 1 , x < 0 x > 0 We see that lim x → 0 + f ( x ) = 1 \lim_{x \to 0^+} f(x) = 1 lim x → 0 + f ( x ) = 1 (the value from the right), while
lim x → 0 − f ( x ) = − 1 \lim_{x \to 0^-} f(x) = -1 lim x → 0 − f ( x ) = − 1 (the value from the left). Thus,
lim x → 0 f ( x ) \lim_{x \to 0} f(x) lim x → 0 f ( x ) does not exist.
Normal limit laws also still apply to one-sided limits:
lim x → − 2 + ( x x + 1 ) ( 2 x + 5 x 2 + x ) = ( − 2 − 2 + 1 ) ( − 4 + 5 4 − 2 ) = 1 \lim_{x \to -2^+} \left( \frac{x}{x+1} \right) \left( \frac{2x + 5}{x^2 + x} \right)
= \left( \frac{-2}{-2 + 1} \right) \left( \frac{-4 + 5}{4 - 2} \right) = 1 x → − 2 + lim ( x + 1 x ) ( x 2 + x 2 x + 5 ) = ( − 2 + 1 − 2 ) ( 4 − 2 − 4 + 5 ) = 1 However, be careful about which piece of the function applies:
lim x → − 2 − ( x + 3 ) ∣ x + 2 ∣ x + 2 = ( lim x → − 2 − x + 3 ) ( lim x → − 2 − ∣ x + 2 ∣ x + 2 ) = ( − 2 + 3 ) ( − 1 ) = − 1 \begin{align}
\lim_{x \to -2^-} (x + 3) \frac{|x + 2|}{x + 2}
&= \left( \lim_{x \to -2^-} x + 3 \right)
\left( \lim_{x \to -2^-} \frac{|x + 2|}{x + 2} \right) \\
&= (-2 + 3)(-1) = -1
\end{align} x → − 2 − lim ( x + 3 ) x + 2 ∣ x + 2∣ = ( x → − 2 − lim x + 3 ) ( x → − 2 − lim x + 2 ∣ x + 2∣ ) = ( − 2 + 3 ) ( − 1 ) = − 1 where we used the limit we already found of f ( x ) = ∣ x ∣ / x f(x) = |x|/x f ( x ) = ∣ x ∣/ x above to evaluate
the second limit.
One-sided limit challenge ¶ Using
Find the limits at x = 1 x = 1 x = 1 and x = 2 x = 2 x = 2 .
From the graph, lim x → 1 − f ( x ) = 0 \lim_{x \to 1^-} f(x) = 0 lim x → 1 − f ( x ) = 0 and lim x → 1 + f ( x ) = 1 \lim_{x \to 1^+} f(x) = 1 lim x → 1 + f ( x ) = 1 , so
lim x → 1 f ( x ) \lim_{x \to 1} f(x) lim x → 1 f ( x ) does not exist. However, lim x → 2 − f ( x ) = 1 \lim_{x \to 2^-} f(x) = 1 lim x → 2 − f ( x ) = 1 and
lim x → 2 + f ( x ) = 1 \lim_{x \to 2^+} f(x) = 1 lim x → 2 + f ( x ) = 1 , so lim x → 2 f ( x ) = 1 \lim_{x \to 2} f(x) = 1 lim x → 2 f ( x ) = 1 , even though f ( 2 ) = 2 f(2) = 2 f ( 2 ) = 2 .
Infinite limits ¶ Some functions tend to ± ∞ \pm\infty ± ∞ at a value, even if they do not have a limit.
f ( x ) = 1 / x 2 f(x) = 1/x^2 f ( x ) = 1/ x 2 Limit exists and is + ∞ +\infty + ∞ .
g ( x ) = 1 / x g(x) = 1/x g ( x ) = 1/ x Limit does not exist.
Infinite limits
f ( x ) f(x) f ( x ) approaches ∞ \infty ∞ as x approaches x 0 x_0 x 0
lim x → ∞ f ( x ) = ∞ \lim_{x\to\infty}f(x) = \infty x → ∞ lim f ( x ) = ∞ if for every real B there exists δ > 0 \delta > 0 δ > 0 such that for all x if
0 < ∣ x − x 0 ∣ < δ 0 < |x - x_0| < \delta 0 < ∣ x − x 0 ∣ < δ then f ( x ) > f(x) > f ( x ) > B. There is an analogous definition
for functions that approach − ∞ -\infty − ∞ .
This behavior can be used to define a vertical asymptote of a function:
L’Hôpital’s Rule ¶ L’Hôpital’s rule is a method that allows you to calculate limits of certain
indeterminate forms using derivatives.
L’Hôpital’s Rule
If lim x → a f ( x ) = lim x → a g ( x ) = 0 \lim_{x\to a} f(x) = \lim_{x \to a} g(x) = 0 lim x → a f ( x ) = lim x → a g ( x ) = 0 or ± ∞ \pm \infty ± ∞ , then
lim x → a f ( x ) g ( x ) = lim x → a f ′ ( x ) g ′ ( x ) \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} x → a lim g ( x ) f ( x ) = x → a lim g ′ ( x ) f ′ ( x ) This process can be repeated until the above situation no longer applies or
the limit is found not to exist.
Let’s work some examples:
The limits of the numerator and denominator are both zero, so:
lim x → 0 sin x x = lim x → 0 cos x 1 = 1 \lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\cos x}{1} = 1 x → 0 lim x sin x = x → 0 lim 1 cos x = 1 The limits of the numerator and denominator are both zero, so:
lim x → 0 x − sin x x 3 = lim x → 0 1 − cos x 3 x 2 = lim x → 0 sin x 6 x = 1 6 lim x → 0 sin x x = 1 6 \begin{align}
\lim_{x\to 0} \frac{x-\sin x}{x^3}
&= \lim_{x\to 0} \frac{1-\cos x}{3x^2} \\
&= \lim_{x\to 0} \frac{\sin x}{6x} \\
&= \frac{1}{6} \lim_{x\to 0} \frac{\sin x}{x} = \frac{1}{6}
\end{align} x → 0 lim x 3 x − sin x = x → 0 lim 3 x 2 1 − cos x = x → 0 lim 6 x sin x = 6 1 x → 0 lim x sin x = 6 1 where we used L’Hôpital’s rule twice, then used the result from the first
example.
The limits of the numerator and denominator are both ∞ \infty ∞ , so:
lim x → ∞ ln x 2 x = lim x → ∞ 1 / x 1 / x = lim x → ∞ 1 x = 0 \begin{align}
\lim_{x\to \infty} \frac{\ln x}{2\sqrt{x}}
&= \lim_{x\to \infty} \frac{1/x}{1/\sqrt{x}} \\
&= \lim_{x\to \infty} \frac{1}{\sqrt{x}} = 0
\end{align} x → ∞ lim 2 x ln x = x → ∞ lim 1/ x 1/ x = x → ∞ lim x 1 = 0 where the last limit is obtained by simplification.
The limits of the numerator and denominator are both ∞ \infty ∞ , so:
lim x → ∞ e x x 2 = lim x → ∞ e x 2 x = lim x → ∞ e x 2 = ∞ \begin{align}
\lim_{x\to \infty} \frac{e^x}{x^2}
&= \lim_{x\to \infty} \frac{e^x}{2x} \\
&= \lim_{x\to \infty} \frac{e^x}{2} = \infty
\end{align} x → ∞ lim x 2 e x = x → ∞ lim 2 x e x = x → ∞ lim 2 e x = ∞ Here, we used L’Hôpital’s rule twice, but then ultimately found that the
limit does not exist.
This limit is not in a form that’s immediately suitable for L’Hôpital’s rule,
but it can be made so.
lim x → ∞ x sin ( 1 / x ) = lim x → ∞ sin ( 1 / x ) 1 / x = lim x → ∞ cos ( 1 / x ) ( − 1 / x 2 ) − 1 / x 2 = lim x → ∞ cos ( 1 / x ) = 1 \begin{align}
\lim_{x\to \infty} x\sin(1/x)
&= \lim_{x\to \infty} \frac{\sin(1/x)}{1/x} \\
&= \lim_{x\to \infty} \frac{\cos(1/x)(-1/x^2)}{-1/x^2} \\
&= \lim_{x\to \infty} \cos(1/x) = 1
\end{align} x → ∞ lim x sin ( 1/ x ) = x → ∞ lim 1/ x sin ( 1/ x ) = x → ∞ lim − 1/ x 2 cos ( 1/ x ) ( − 1/ x 2 ) = x → ∞ lim cos ( 1/ x ) = 1 Another option in this case would be to substitute in y = 1 / x y = 1/x y = 1/ x , then note
that y → 0 y \to 0 y → 0 as x → ∞ x \to \infty x → ∞ so we have
lim y → 0 sin y y = 1 \lim_{y\to 0} \frac{\sin y}{y} = 1 y → 0 lim y sin y = 1 which we already solved!