Limit Formula

If a given function y = f(x) is not defined at the point x = a. but the function can be defined at neighborhood values of x = a. That’s the limiting behavior of a function. limit helps us to determine the values of a function at x = a, but not the exact value, it’s an approaching value of x at a. When x approaches a and we can write as lim_x⇢a f(x). The approaching and exact value has a very very small difference between them i.e. if the exact point is 2 then the approaching value is 1.9999999… so on. This difference can’t affect any major.

Most students believe that a limit should be a finite number. But it is quite possible that f(x) had an infinite limit as x⇢a. i.e.  lim_x⇢a f(x) = ∞.

Limit Formulae

Trigonometric limits: To evaluate trigonometric limits, we have to reduce the terms of the function into simpler terms or into terms of sinθ and cosθ.  

• lim_{x ⇢ 0} [Tex]\frac{sinx}{x} [/Tex] = lim_{x ⇢ 0} [Tex]\frac{x}{sinx} [/Tex] = 1
• lim_{x ⇢ 0} tanx/x = lim_{x ⇢ 0} x/tanx =1

As we considered our first one, 

lim_{x ⇢ 0} sinx/x =1      

Using L-Hospital

lim_{x ⇢ 0} cosx/1 

lim_{x ⇢ 0} cos(0)/1 = 1/1 =1

If the function gives an indeterminate form by putting limits, Then use the l-hospital rule.

Indeterminate  Form 

0/0, ∞/∞, ∞-∞, ∞/0, 0^∞, ∞⁰ , 0⁰, ∞^∞

L-hospital Rule

If we get the indeterminate form, then we differentiate the numerator and denominator separately until we get a finite value. Remember we would differentiate the numerator and denominator the same number of times. Similarly for all trigonometric function,

• lim_{x ⇢ 0} sin^-1x/x = lim_{x ⇢ 0} x/sin^-1x = 1

lim_{x ⇢ 0} sin^-1x/x =1

lim_{x ⇢ 0} 1/√1+x²  [Using L-Hospital] 

= 1/√(1 + (0)²) = 1

• lim_x⇢0[Tex] \frac{tan^{-1}x}{x}  [/Tex]=1
• lim_{x ⇢ a}  sin x^o/x = π/180
• lim_x⇢0 cosx = 1
• lim_{x ⇢ a}  sin(x-a) / (x-a) =1

lim_{x ⇢ a} sin(x – a) / (x – a) 

=1

lim_{x ⇢ a} cos(x – a)/1 

= lim_{x ⇢ a} cos(a – a) = cos(0) =1

• lim_x⇢∞ sinx/x = 0
• lim_x⇢∞ cosx/x = 0
• lim_x⇢∞ sin(1/x) / (1/x) =0

lim_{x ⇢ ∞} sin(1/x)/(1/x) = 0

Let 1/x = h

So, limits changes to 0

Because 1\∞ = 0

lim_{h ⇢ 0} sinh/h

As we see before, If lim_{x ⇢ 0} sinx/x = 1

So, lim_{h ⇢ 0} sinh/h = 1

Exponential limits 

• lim_{x ⇢ 0}  e^x – 1 /x = 1
• lim_{x ⇢ 0}  a^x – 1 /x = log_ea
• lim_{x ⇢ 0} e^λx – 1 /x = λ

Same here, we get our desired result by using L-hospital rule. 

Alternate method: Using expansion

e^x = 1 + X + X²/2! + X³/3! + X⁴/4!+ … ∞

lim_{x ⇢ 0}  e^x – 1 /x = 1

lim_{x ⇢ 0} (1 + X + X²/2!+ —) -1 /x

lim_{x ⇢ 0} (X + X²/2! + —)/x

lim_{x ⇢ 0} 1 + X + X²/2!+—

lim_{x ⇢ 0} 1 + 0 + 0 + 0 + 0— = 1

Logarithmic limits

• lim_{x ⇢ 0} log(1 + x) /x = 1
• lim_{x ⇢ e} log_ex = 1
• lim_{x ⇢ 0} log_e(1 – x) /x  = -1
• lim_{x ⇢ 0} log_a(1 + x) /x = log_ae

Simply proved by using L-hospital and expansion method.

Some Important expansion 

[Tex]e^x = 1+\frac{x}{1!} +\frac{x^2}{2!} + \frac{x^3}{3!} +— [/Tex]
[Tex]a^x = 1+ xloga + \frac{x^2(loga)^2}{2!} + \frac{x^3(loga)^3}{3!} +— [/Tex]
[Tex]log(1+x) = x- \frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4} +— [/Tex]
[Tex]log(1-x) = -x- \frac{x^2}{2} – \frac{x^3}{3} -\frac{x^4}{4} +— [/Tex]
[Tex](1+x)n = 1 + nx + \frac{n(n-1)x^2}{2!} +— [/Tex]
[Tex]sinx = x- \frac{x^3}{3!} + \frac{x^5}{5!} +— [/Tex]
[Tex]cosx = 1- \frac{x^2}{2!} + \frac{x^4}{4!} +— [/Tex]
[Tex]tanx = x + \frac{x^3}{3} + \frac{2x^5}{15} +— [/Tex]
[Tex]sin^{-1}x = x + \frac{x^3}{3!} + \frac{9x^5}{5!} +— [/Tex]
[Tex]cos^{-1}x = x – \frac{x^3}{6} + — [/Tex]
[Tex]tan^{-1}x= x – \frac{x^3}{3} + \frac{x^5}{5} +— [/Tex]
[Tex]sinhx =  x+ \frac{x^3}{3!} + \frac{x^5}{5!} +—        [/Tex] (Here, sinhx is a hyperbolic function)
[Tex]coshx = 1+ \frac{x^2}{2!} + \frac{x^4}{4!} +— [/Tex]
[Tex]tanhx = x – \frac{x^3}{3} + \frac{2x^5}{15} +— [/Tex]
[Tex](1+x)\frac{1}{x} = e^{1- \frac{x}{2} + \frac{11×4}{24} } +— [/Tex]

Important Results : 

1. lim_x⇢0 [Tex](1+x)^{\frac{1}{x}}  [/Tex]= e
2. lim_x⇢0 [Tex]\left( 1+\frac{1}{x} \right)^x [/Tex] = e
3. lim_x⇢0 [Tex]\frac{e^x-1}{x} =1 [/Tex]
4. lim_x⇢0[Tex]\frac{a^x-1}{x} [/Tex] = log_ea
5. lim_x⇢0[Tex] \frac{1-cosmx}{x^2} [/Tex] = m²/2
6. lim_x⇢0 [Tex]\frac{1-cosmx}{1-cosnx} = \frac{m^2}{n^2} [/Tex]

Shortcut : 

1. lim_x⇢0 [Tex]\left( 1+ \frac{a}{b}x \right)^\frac{c}{dx} = e^\frac{ac}{bd} [/Tex]
2. lim_x⇢0[Tex] \left( 1+ \frac{a}{bx} \right)^\frac{cx}{d} = e^\frac{ac}{bd} [/Tex]
3. lim_x⇢a f(x)^g(x) = e^{limx\to a\left[ f(x)-1 \right].g(x)}

Sample problem

Question 1: Solve, lim_x⇢0  (x – sinx ) /(1 – cosx).

Solution:

Using L-hospital,

lim_{x ⇢ 0} (1 – cosx) / (sinx)

lim_{x ⇢ 0} sinx / cosx = sin(0) / cos(0) = 0/1 = 0

Question 2: Solve, lim_{x ⇢ 0} (e^2x -1) / sin4x.

Solution:

Using L-hospital 

lim_{x ⇢ 0} (2)(e^2x) / cos4x

lim_{x ⇢ 0}  2(e⁰) / cos4(0) = 2/1= 2

Question 3: Solve, lim_{x ⇢ 0} (1 – cosx) / x²

Solution:

Using L-hospital 

lim_{x ⇢ 0} sinx /2x = 1/2 {sinx/x = 1}

Question 4: Solve, lim_{x ⇢ ∞} [Tex]\frac{x+sinx}{x} [/Tex]

Solution:

lim_{x ⇢ ∞} (1 + [Tex]\frac{sinx}{x}        [/Tex])

1 + lim_{x ⇢ ∞}[Tex] \frac{sinx}{x} [/Tex]

As we know, x = ∞  

So 1/x = 0

[Tex]1 + lim\frac{1}{x}⇢∞ \frac{sin\frac{1}{x}}{x} [/Tex]

1 + 0 = 0                                                                                                                                                 

Question 5: Solve, lim_{x ⇢ π/2} (tanx)^cosx 

Solution:

let Y = lim_{x ⇢ π/2}  (tanx)^cosx

Taking log_e both sides,

 log_eY = lim_{x ⇢ π/2}  log_e(tanx)^cosx 

 log_eY = lim_{x ⇢ π/2}  cosx log_e(tanx)

log_ey = lim_{x ⇢ π/2} loge(tanx)/secx

Using l-hospital,

log_ey = lim_{x ⇢ π/2} cosx /sin²x = 0

Now, taking exponent on both sides,

Y = lim_{x ⇢ π/2} e⁰

Y = lim_{x ⇢ π/2}  (tanx)^cosx = 1  

Question 6: lim_{x ⇢ 0} [Tex]\frac {e^x -(1+x+ \frac {x^2}{2})}{ x^3} [/Tex]

Solution:

limx⇢0 \frac{1+\frac{x}{1!} + \frac{x2}{2!} + \frac{x3}{3!} – ( 1+ x+ \frac{x2}{2!} ) }{x3}

limx⇢0 [Tex]\frac{\frac{x3}{3!}}{x3}  [/Tex]= 1/3! =1/6

Question 7: Solve, lim_{a ⇢ 0}  [Tex]\frac{x^a-1}{a} [/Tex]

Solution:

Using l-hospital (Differentiating numerator and denominator w.r.t a)  

lim_{a ⇢ 0}  x^alogx = logx

Question 8: Solve, lim_{x ⇢ 0} [Tex]\frac{x^2+x-sinx}{x^2} [/Tex]

Solution:

lim_{x ⇢ 0} [Tex]\frac{x^2+x-(x-x^3/3!)}{x^2} [/Tex]

lim_{x ⇢ 0} 1 + x/3! = 1