English
Tamil Nadu Board of Secondary EducationHSC Science Class 11

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why? limx→31x-3

Advertisements
Advertisements

Question

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> 3) 1/(x - 3)`

Graph
Advertisements

Solution

To find `lim_(x -> 3) 1/(x - 3)`

y = f(x) = `1/(x - 3)`

From the graph the value of the function at x = 3 the curve does not meet the line x = 3

∴ The value of the function is not defined at the point x = 3.

Hence `lim_(x -> 3) 1/(x - 3)` does not exist at x = 3.

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [Page 96]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 11 TN Board
Chapter 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 11 | Page 96

RELATED QUESTIONS

Evaluate the following limit:

`lim_(x -> 3)[sqrt(2x + 6)/x]`


Evaluate the following limit:

`lim_(x -> 5)[(x^3 - 125)/(x^5 - 3125)]`


Evaluate the following limit :

`lim_(x -> 1)[(x + x^2 + x^3 + ......... + x^"n" - "n")/(x - 1)]`


Evaluate the following :

Given that 7x ≤ f(x) ≤ 3x2 – 6 for all x. Determine the value of `lim_(x -> 3) "f"(x)`


Evaluate the following :

`lim_(x -> 1) [(x + 3x^2 + 5x^3 + ... + (2"n" - 1)x^"n" - "n"^2)/(x - 1)]`


In problems 1 – 6, using the table estimate the value of the limit.

`lim_(x -> 2) (x - 2)/(x^2 - x - 2)`

x 1.9 1.99 1.999 2.001 2.01 2.1
f(x) 0.344820 0.33444 0.33344 0.333222 0.33222 0.332258

In exercise problems 7 – 15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

`lim_(x -> 2) f(x)` where `f(x) = {{:(4 - x",", x ≠ 2),(0",", x = 2):}`


Evaluate the following limits:

`lim_(sqrt(x) -> 3) (x^2 - 81)/(sqrt(x) - 3)`


Evaluate the following limits:

`lim_(x -> 3) (x^2 - 9)/(x^2(x^2 - 6x + 9))`


Evaluate the following limits:

`lim_(x  -> oo) 3/(x - 2) - (2x + 11)/(x^2 + x - 6)`


Evaluate the following limits:

`lim_(x -> oo) (1 + 3/x)^(x + 2)`


Evaluate the following limits:

`lim_(alpha -> 0) (sin(alpha^"n"))/(sin alpha)^"m"`


Evaluate the following limits:

`lim_(x -> 0) (sin("a" + x) - sin("a" - x))/x`


Evaluate the following limits:

`lim_(x -> 0) (3^x - 1)/(sqrt(x + 1) - 1)`


Evaluate the following limits:

`lim_(x -> 0) ("e"^("a"x) - "e"^("b"x))/x`


Choose the correct alternative:

`lim_(x - oo) sqrt(x^2 - 1)/(2x + 1)` =


If `lim_(x -> 1) (x + x^2 + x^3|+ .... + x^n - n)/(x - 1)` = 820, (n ∈ N) then the value of n is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×