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}  = lim_{x ⇢ 0}  = 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=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 

 (Here, sinhx is a hyperbolic function)

Important Results : 

1. lim_x⇢0 = e
2. lim_x⇢0  = e
3. lim_x⇢0
4. lim_x⇢0 = log_ea
5. lim_x⇢0 = m²/2
6. lim_x⇢0 

Shortcut : 

1. lim_x⇢0
2. lim_x⇢0
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 ⇢ ∞}

Solution:

lim_{x ⇢ ∞} (1 + )

1 + lim_{x ⇢ ∞}

As we know, x = ∞  

So 1/x = 0

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}

Solution:

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

limx⇢0 = 1/3! =1/6

Question 7: Solve, lim_{a ⇢ 0} 

Solution:

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

lim_{a ⇢ 0}  x^alogx = logx

Question 8: Solve, lim_{x ⇢ 0}

Solution:

lim_{x ⇢ 0}

lim_{x ⇢ 0} 1 + x/3! = 1