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Finding Limits Graphically

22nd December is National Mathematics Day celebrated as a birth anniversary of Indian Mathematician Srinivasa Ramanujan so I am Anita Govilkar the owner of the website www.ask-math.com dedicate this page to a great mathematician.

finding limits graphically is find out by visualizing the graph.
Let us consider that the limit of f(x) as x approaches a is equal to L, written

$\lim_{x->a}f(x) =L $

if we can make the values of f(x) from the graph as close to L as we like by taking x from the graph to be sufficiently close to a, but not equal to a. In other words we can say that, as x approaches a (but not equal to a), f(x) approaches L.
So in this case we can say that the limit of f(x) as x approaches a from the left side is equal to L, written

$\lim_{x->a^{-}}f(x) = L $

The negative exponent represents that the x approaches to a from the left side.
And the limit of f(x) as x approaches a from the right side is equal to L , written

$\lim_{x->a^{+}}f(x) = L $

When the value of left side limit and right side limit are not equal then the limit of that function does not exists.

$\lim_{x->a^{-}}f(x) \neq \lim_{x->a^{+}}f(x)$
From the above graph we can easily say that as x approaches to 1 from left side the limit approaches to 2 and when x approaches to 1 from right side the limit approaches to 2. This can be written as follows

$ \lim_{x->1^{-}}f(x)= 2$ and $\lim_{x->1^{+}}f(x) = 2 $

Examples on Finding Limits Graphically

Example 1 : Use the following graph find the limit if exists.
Solution :
1) In the diagram 1, we can see that as x approaches 3 from the left side the value of function approaches to 1 and x approaches 3 from the right side the value of function approaches 1.
we approach the same value from both sides then the limit is equal to 1.

$\lim_{x->3^{-}}(4-x) = 1$ and $\lim_{x->3^{+}}(4-x) = 1$
So the given limit exists and its value is 1.

2) In the diagram 2, we can see that as x approaches 0 from the left side the value of function approaches to 1 and x approaches 0 from the right side the value of function approaches 1.
we approach the same value from both sides then the limit is equal to 1.

$\lim_{x->0^{-}}sec(x) = 1$ and $\lim_{x->0^{+}}sec(x) = 1$
So the given limit exists and its value is 1.

3) In the diagram 3, we can see that as x approaches 2 from the left side the value of function approaches to -1 and x approaches 2 from the right side the value of function approaches +1.

$\lim_{x->2^{-}}\frac{|x-2|}{x-2} = -1$ and $\lim_{x->2^{+}}\frac{|x-2|}{x-2} = 1$
Since we approach the two different values from both sides so the limit does not exist.

4) In the diagram 4, we can see that as x approaches 5 from the left side the value of function approaches to $- \infty $ and x approaches 5 from the right side the value of function approaches $+ \infty $.

$\lim_{x->5^{-}}\frac{2}{x-5} = - \infty$ and $\lim_{x->5^{+}}\frac{2}{x-5} = + \infty$
Since we approach the two different values from both sides so the limit does not exist.



12th grade math

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