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Explain :- limit h tends to 0 hsin1/h = 0?
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Explain :- limit h tends to 0 hsin1/h = 0?
Simple technique anynumber multiplied with zero gives zero By direct substitution method
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Explain :- limit h tends to 0 hsin1/h = 0?
Explanation:


To find the limit of the expression as h approaches 0, we can use the concept of trigonometric limits. Let's break down the expression step by step to understand how the limit evaluates to 0.

Step 1: Simplifying the Expression


The given expression is hsin(1/h). To simplify it further, we can rewrite it as sin(1/h)/h. This step allows us to work with a more manageable form of the expression.

Step 2: Applying Trigonometric Limits


To evaluate the limit, we need to determine the behavior of sin(1/h)/h as h approaches 0. Trigonometric limits tell us that the limit of sin(x)/x as x approaches 0 is equal to 1.

Key Point:

Trigonometric Limit: lim(x→0) sin(x)/x = 1

In our expression, we have sin(1/h)/h. By substituting x = 1/h, we can rewrite the expression as sin(x)/x. Therefore, the limit of hsin(1/h) as h approaches 0 is equivalent to sin(x)/x as x approaches infinity.

Step 3: Evaluating the Limit


Using the trigonometric limit mentioned above, we can conclude that sin(x)/x approaches 1 as x approaches 0. Since h approaches 0, we can substitute x = 1/h back into the expression.

Therefore, sin(x)/x = sin(1/h)/h approaches 1 as h approaches 0.

Key Point:

Trigonometric Limit: lim(x→0) sin(x)/x = 1

Step 4: Final Result


By applying the trigonometric limit, we have established that the limit of hsin(1/h) as h approaches 0 is equal to 1. Therefore, the given expression, hsin(1/h), evaluates to 0 as h tends to 0.

Key Point:

The limit of hsin(1/h) as h approaches 0 is 0.

Note: It is important to note that this limit does not imply that hsin(1/h) is equal to 0 for all values of h. The limit only gives us information about the behavior of the expression as h approaches 0.
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Explain :- limit h tends to 0 hsin1/h = 0?
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