How do you find the limit of #sin(x^2)/x# as #theta->0#?

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How do you find the limit of #sin(x^2)/x# as #x->0#?

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Answer 1 · 2017-01-16T13:54:06

#" Reqd. Lim."=0.#

Reqd. Lim. #=lim_(xrarr0)sin(x^2)/x#

#=lim_(xrarr0) (sinx^2/x^2)(x^2/x)#

#=lim_(xrarr0)(sinx^2/x^2)(x)#

#={lim_(thetararr0)sintheta/theta}{lim_(xrarr0) x}#..........

[#"where, "theta=x^2," and, as "xrarr0, theta rarr0.]#

#=(1)(0)#.

#:." Reqd. Lim."=0.#